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Linear algebra: the theory of vector spaces and linear transformations

Aaron Greicius

Section 5.2 Orthogonal bases

Subsection 5.2.1 Orthogonal vectors and sets

Definition 5.2.1. Orthogonality.

Let \((V,\langle \ , \rangle)\) be an inner product space. Vectors \(\boldv, \boldw\in V\) are orthogonal if \(\langle \boldv, \boldw\rangle =0\text{.}\)
Let \(S\subseteq V\) be a set of nonzero vectors.
  • The set \(S\) is orthogonal if \(\langle\boldv,\boldw \rangle=0\) for all \(\boldv\ne\boldw\in S\text{.}\) We say that the elements of \(S\) are pairwise orthogonal in this case.
  • The set \(S\) is orthonormal if it is both orthogonal and satisfies \(\norm{\boldv}=1\) for all \(\boldv\in S\text{:}\) i.e., \(S\) consists of pairwise orthogonal unit vectors.
Given any distinct elements \(\boldv_1, \boldv_2, \dots, \boldv_r\in S\text{,}\) we have
\begin{align*} c_1\boldv_1 +c_2\boldv_2+\cdots +c_r\boldv_r=\boldzero\amp \Rightarrow\amp \langle c_1\boldv_1 +c_2\boldv_2 +\cdots +c_r\boldv_r,\boldv_i\rangle=\langle\boldzero,\boldv_i\rangle\\ \amp \Rightarrow\amp c_1\langle\boldv_1,\boldv_i\rangle +c_2\langle \boldv_2,\boldv_i\rangle +\cdots +c_r\langle\boldv_r,\boldv_i\rangle=0\\ \amp \Rightarrow\amp c_i\langle \boldv_i,\boldv_i\rangle=0 \ \text{ (since \(\langle\boldv_j,\boldv_i\rangle= 0\) for \(j\ne i\)) }\\ \amp \Rightarrow\amp c_i=0 \text{ (since \(\langle\boldv_i,\boldv_i\rangle\ne 0\)) }\text{.} \end{align*}
This proves that \(S\) is linearly independent.

Example 5.2.3.

Show that the set \(S=\{\boldx_1=(1,1,1),\boldx_2=(1,2,-3), \boldx_3=(-5,4,1)\}\) is orthogonal with respect to the dot product. Explain why it follows that \(S\) is a basis of \(\R^3\text{.}\)
Solution.
A simple computation shows \(\boldx_i\cdot \boldx_j=0\) for all \(1\leq i\ne j\leq 3\text{,}\) showing that \(S\) is orthogonal. Theorem 5.2.2 implies \(S\) is linearly independent. Since \(\val{S}=\dim \R^3=3\text{,}\) it follows from Corollary 3.7.12 that \(S\) is a basis.

Example 5.2.4.

Let \(V=C([0,2\pi])\) with integral inner product \(\langle f, g\rangle=\int_0^{2\pi} f(x)g(x) \, dx\text{,}\) and let
\begin{equation*} S=\{1, \cos(x),\sin(x),\cos(2x),\sin(2x), \dots\}=\{\cos(nx)\colon n\in\Z_{>0}\}\cup\{\sin(mx)\colon m\in\Z_{>0}\}\text{,} \end{equation*}
where the element \(1\in S\) is understood as the constant function \(f(x)=1\) for all \(x\in [0,2\pi]\text{.}\) Show that \(S\) is orthogonal and hence linearly independent.
Solution.
First observe that
\begin{align*} \angvec{1,1} \amp = \int_0^{2\pi} 1\, dx=2\pi \\ \angvec{1, \cos n x} \amp= \int_0^{2\pi}\cos n x\, dx=0 \\ \angvec{1, \sin n x} \amp= \int_0^{2\pi}\sin n x\, dx=0 \end{align*}
for all \(n\text{.}\) (Note: since \(\angvec{1,1}=2\pi\ne 1\text{,}\) the set \(S\) is not orthonormal. ) Next, using the trigonometric identities
\begin{align*} \cos\alpha\cos\beta \amp =\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))\\ \sin\alpha\sin\beta \amp=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta)) \\ \sin\alpha\cos\beta \amp =\frac{1}{2}(\sin(\alpha-\beta)+\sin(\alpha+\beta)) \end{align*}
it follows that
\begin{align*} \langle \cos(nx),\cos(mx)\rangle=\int_0^{2\pi}\cos(nx)\cos(mx)\, dx\amp =\begin{cases} 0\amp \text{ if \(n\ne m\) }\\ \pi\amp \text{ if \(n=m\) } \end{cases}\\ \langle \sin(nx),\sin(mx)\rangle=\int_0^{2\pi}\sin(nx)\sin(mx)\, dx\amp =\begin{cases} 0\amp \text{ if \(n\ne m\) }\\ \pi\amp \text{ if \(n=m\) } \end{cases}\\ \langle \cos(nx),\sin(mx)\rangle=\int_0^{2\pi}\cos(nx)\sin(mx)\, dx\amp =0 \text{ for any \(n,m\) }\text{.} \end{align*}

Subsection 5.2.2 Orthogonal bases

Given an inner product space \(V\) we will see that working with orthogonal sets of vectors is extremely convenient computationally speaking. In particular, when picking basis of \(V\text{,}\) we will look for one consisting of orthogonal vectors. Not surprisingly, this is called an orthogonal basis.

Definition 5.2.5. Orthogonal and orthonormal bases.

Let \((V,\langle \ , \rangle)\) be an inner product space. An orthogonal basis (resp., orthonormal basis) of \(V\) is a basis \(B\) that is orthogonal (resp., orthonormal) as a set.
We will see in Theorem 5.2.8 precisely why working with orthogonal or orthonormal bases is so convenient. Before we do so, however, we would like some guarantee that we can actually find an orthogonal basis. The Gram-Schmidt procedure comes to our rescue in this regard, at least in the finite-dimensional case, as it provides a method of converting an arbitrary basis into an orthogonal one.
  1. See Procedure 5.2.6 and its proof.
  2. The orthogonal set \(S\) is linearly independent by Theorem 5.2.2. Let \(S=\{\boldv_1,\boldv_2,\dots, \boldv_r\}\) be the distinct elements of \(S\text{.}\) (We must have \(r\leq n\) since \(S\) is linearly independent.) By Theorem 3.7.11 we can extend \(S\) to a basis \(B=\{\boldv_1,\dots, \boldv_r, \boldv_{r+1}, \dots, \boldv_n\}\text{.}\) It is easy to see that when we apply the Gram-Schmidt procedure to \(B\text{,}\) the first \(r\) vectors are left unchanged, as they are already pairwise orthogonal. Thus Gram-Schmidt returns an orthogonal basis of the form \(B'=\{\boldv_1,\dots, \boldv_r, \boldw_{r+1}, \dots, \boldw_n\}\text{,}\) as desired.
Now let’s see the computational virtue of working with orthogonal bases.

Example 5.2.9.

Consider the inner product space \(V=\R^2\) with the dot product.
  1. Verify that \(B'=\{\boldv_1=(\sqrt{3}/2,1/2), \boldv_2=(-1/2,\sqrt{3}/2)\}\) is an orthonormal basis of \(\R^2\text{.}\)
  2. Let \(\boldv=(4,2)\text{.}\) Find the scalars \(c_1, c_2\in \R\) such that \(\boldv=c_1\boldv_1+c_2\boldv_2\text{.}\)
  3. Verify that \(\norm{\boldv}=\sqrt{c_1^2+c_2^2}\text{.}\)
Solution.
  1. Easily verified.
  2. Using Theorem 5.2.8 we compute
    \begin{align*} c_1 \amp =\boldv\cdot \boldv_1=2\sqrt{3}+1 \\ c_2\amp= \boldv\cdot \boldv_2=\sqrt{3}-2 \text{.} \end{align*}
  3. Computing directly yields \(\norm{\boldv}=\sqrt{20}=2\sqrt{5}\text{.}\) Using the generalized Pythagorean theorem we have
    \begin{align*} \norm{\boldv} \amp= \sqrt{(2\sqrt{3}+1)^2+(\sqrt{3}-2)^2} \\ \amp=\sqrt{(12+4\sqrt{3}+1)+(3-4\sqrt{3}+4)} \\ \amp = \sqrt{20}=2\sqrt{5}\text{,} \end{align*}
    as desired.
As the previous example and Theorem 5.2.8 begin to make clear, orthogonal bases, and especially orthonormal bases make our life easier computationally speaking. This observation is worthy of a mantra.

Subsection 5.2.3 Coordinate vectors and matrix representations

Let \(V\) be an inner product space. By an orthogonal (resp., orthonormal) ordered basis of \(V\text{,}\) we mean an ordered basis \((\boldv_1, \boldv_2, \dots, \boldv_n)\) for which the underlying set \(\{\boldv_1, \boldv_2, \dots, \boldv_n\}\) is orthogonal (resp., orthonormal). It should come as little surprise that as a consequence of Theorem 5.2.8, computing coordinate vectors and matrix representations with respect to orthogonal bases is especially easy.
For example, if \(B=(\boldv_1, \boldv_2, \dots, \boldv_n)\) is an orthogonal basis with respect to some inner product \(\langle , \rangle \text{,}\) then by (5.2.1) we have
\begin{equation*} \boldv=\frac{\langle \boldv,\boldv_1 \rangle }{\langle \boldv_1, \boldv_1\rangle }\boldv_1+\frac{\langle \boldv,\boldv_1 \rangle }{\langle \boldv_1, \boldv_1\rangle }\boldv_1+\cdots +\frac{\langle \boldv,\boldv_n \rangle }{\langle \boldv_n, \boldv_n\rangle }\boldv_n\text{,} \end{equation*}
and thus, using definition Definition 4.1.3, we have
\begin{equation*} [\boldv]_B=\left(\frac{\langle \boldv,\boldv_1 \rangle }{\langle \boldv_1, \boldv_1\rangle }, \frac{\langle \boldv,\boldv_2 \rangle }{\langle \boldv_2, \boldv_2\rangle },\dots, \frac{\langle \boldv,\boldv_n \rangle }{\langle \boldv_n, \boldv_n\rangle }\right)\text{.} \end{equation*}
We have thus proved the following theorem.

Example 5.2.12. Orthogonal bases.

Let \(V=\R^2\) and \(B=((1,1),(-1,2))\text{.}\) Find a general formula for \([(a,b)]_B\text{.}\) Note: \(B\) is orthogonal with respect to the weighted dot product
\begin{equation*} \langle (x_1,x_2), (y_1,y_2)\rangle =2x_1y_1+x_2y_2\text{.} \end{equation*}
Solution.
Applying the inner product formula (5.2.3) to \(B\) and the dot product with weights \(2, 1\text{,}\) for any \(\boldv=(a,b)\) we compute
\begin{align*} [(a,b)]_B \amp =\left(\frac{\langle (a,b), (1,1)\rangle }{\langle (1,1),(1,1) \rangle }, \frac{\langle (a,b), (-1,2)\rangle }{\langle (-1,2),(-1,2) \rangle }\right)\\ \amp=\left(\frac{1}{3}(2a+b),\frac{1}{3}(-a+b) \right) \text{.} \end{align*}
Let’s check our formula with \(\boldv=(3,-3)\text{.}\) The formula yields \([(3,-3)]_B=(1,-2)\text{,}\) and indeed we see that
\begin{equation*} (3,-3)=1(1,1)-2(-1,2)\text{.} \end{equation*}
Similarly, the computations involved in change of basis matrices are significantly easier when at least one of the bases involved is orthogonal. This is a direct consequence of the fact that the change of basis matrix formula (4.3.1) involves computing coordinate vectors; and this is easy to do when the basis in question is orthogonal, thanks to the inner product formula (5.2.3).
Things get even easier if both bases \(B\) and \(B'\) involved are in fact orthonormal. As we show below, it turns out in this case that the inverse of the change of basis matrix is just its transpose:
\begin{equation*} \left(\underset{B\rightarrow B'}{P}\right)^{-1}=\left(\underset{B\rightarrow B'}{P}\right)^T\text{.} \end{equation*}
It follows from (4.3.4) that when both bases are orthonormal, we can obtain one change of basis matrix from the other simply by computing the transpose. An invertible matrix whose inverse is equal to its transpose is called an elementary matrix. We develop some general theory about this special family of matrices (interesting in its own right) before treating change of basis matrices.

Definition 5.2.13. Orthogonal matrices.

An invertible \(n\times n\) matrix \(A\) is orthogonal if \(A^{-1}=A^T\text{.}\)

Remark 5.2.14.

Since for an invertible matrix \(A\) we have \((A^T)^{-1}=(A^{-1})^T\) it follows immediately from Definition 5.2.13 that
\begin{equation*} A \text{ is orthogonal}\iff A^T \text{ is orthogonal} \iff A^{-1} \text{ is orthogonal}\text{.} \end{equation*}

Example 5.2.15.

Let
\begin{equation*} A=\begin{amatrix}[rr]\frac{\sqrt{2}}{2}\amp -\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2} \amp \frac{\sqrt{2}}{2} \end{amatrix}\text{.} \end{equation*}
Check for yourself that \(A^TA=I\text{.}\) Thus \(A\) is an orthogonal matrix.
Now observe that the columns of \(A\) are orthonormal with respect to the dot product:
\begin{align*} \frac{2}{2}(1,1)\cdot \frac{2}{2}(1,1) \amp =1 \amp \frac{2}{2}(1,-1)\cdot \frac{2}{2}(1,-1)\amp=1 \\ \frac{2}{2}(1,1)\cdot \frac{2}{2}(1,-1) \amp=0 \text{.} \end{align*}
This is not a coincidence!
Let \(\boldr_i\) and \(\boldc_i\) be the \(i\)-th row and column of \(A\text{,}\) respectively, for each \(1\leq i\leq n\text{.}\) From Theorem 2.1.22 we see that
\begin{align} A^TA \amp=[\boldc_i\cdot \boldc_j]_{1\leq i,j\leq n} \tag{5.2.5}\\ AA^T \amp=[\boldr_i\cdot \boldr_j]_{1\leq i,j\leq n} \text{.}\tag{5.2.6} \end{align}
We use here that rows of \(A^T\) are the columns of \(A\text{,}\) and the columns of \(A^T\) are the rows of \(A\text{.}\) From (5.2.5)–(5.2.6) it follows easily that
\begin{align*} A^{-1}=A^T \amp \iff A^TA=I \\ \amp \iff \boldc_i\cdot\boldc_j=\begin{cases} 1\amp \text{if } i=j\\ 0\amp \text{if } i\ne j\end{cases} \\ \amp \iff \{\boldc_1,\boldc_2,\dots, \boldc_n\} \text{ is orthonormal}\\ \amp \iff \{\boldc_1,\boldc_2,\dots, \boldc_n\} \text{ is an orthonormal basis}\amp (n=\dim\R^n)\text{,} \end{align*}
and
\begin{align*} A^{-1}=A^T \amp \iff AA^T=I \\ \amp \iff \boldr_i\cdot\boldr_j=\begin{cases} 1\amp \text{if } i=j\\ 0\amp \text{if } i\ne j\end{cases} \\ \amp \iff \{\boldr_1,\boldr_2,\dots, \boldr_n\} \text{ is orthonormal}\\ \amp \iff \{\boldr_1,\boldr_2,\dots, \boldr_n\} \text{ is an orthonormal basis} \amp (n=\dim\R^n)\text{.} \end{align*}
This proves \((1)\iff (2)\) and \((1)\iff (3)\text{.}\) The result follows.

Remark 5.2.17.

It is somewhat unfortunate that the property of being an orthogonal matrix is equivalent to your rows or columns forming an orthonormal basis. You ask: Why not simply call such matrices orthonormal matrices? My answer: tradition!
  1. Let \(B=(\boldv_1, \boldv_2,\dots, \boldv_n)\text{.}\) By definition, the columns of \(\underset{B\rightarrow B'}{P}\) are the coordinate vectors \([\boldv_i]_{B'}\text{,}\) \(1\leq i\leq n\text{.}\) By Exercise 5.2.4.20, these coordinate vectors form an orthonormal subset of \(\R^n\text{;}\) since there are \(n=\dim\R^n\) of them, they form an orthonormal basis. From Theorem 5.2.16 it follows that \(\underset{B\rightarrow B'}{P}\) is orthogonal. Lastly, from Remark 5.2.14 it follows that \(\underset{B'\rightarrow B}{P}=\left(\underset{B\rightarrow B'}{P}\right)^{-1}\) is also orthogonal.
  2. Since \(\underset{B\rightarrow B'}{P}\) is orthogonal, we have
    \begin{equation*} \underset{B'\rightarrow B}{P}=\left(\underset{B\rightarrow B'}{P}\right)^{-1}=\left(\underset{B\rightarrow B'}{P}\right)^T\text{.} \end{equation*}

Example 5.2.19. Orthonormal change of basis: \(\R^n\).

Let \(B\) be the standard ordered basis of \(\R^3\text{.}\) The ordered basis
\begin{equation*} B'=\left(\frac{1}{\sqrt{6}}(1,1,-2), \frac{1}{\sqrt{2}}(1,-1,0), \frac{1}{\sqrt{3}}(1,1,1) \right) \end{equation*}
is orthonormal with respect to the dot product. Compute the change of basis matrix \(\underset{B\rightarrow B'}{P}\text{.}\)
Solution.
Since \(B\) is the standard ordered basis, we can easily compute
\begin{equation*} \underset{B'\rightarrow B}{P}=\begin{amatrix}[rrr] 1/\sqrt{6}\amp 1/\sqrt{2}\amp 1/\sqrt{3} \\ 1/\sqrt{6} \amp -1/\sqrt{2} \amp 1/\sqrt{3} \\ -2/\sqrt{6} \amp 0 \amp 1/\sqrt{3} \end{amatrix} \text{.} \end{equation*}
Since \(B\) and \(B'\) are both orthonormal bases (with respect to the dot product), we have
\begin{align*} \underset{B\rightarrow B'}{P} \amp = \underset{B'\rightarrow B}{P}^T\\ \amp = \left( \begin{amatrix}[rrr] 1/\sqrt{6}\amp 1/\sqrt{2}\amp 1/\sqrt{3} \\ 1/\sqrt{6} \amp -1/\sqrt{2} \amp 1/\sqrt{3} \\ -2/\sqrt{6} \amp 0 \amp 1/\sqrt{3} \end{amatrix}\right) ^T\\ \amp = \begin{amatrix}[rrr] 1/\sqrt{6}\amp 1/\sqrt{6}\amp -2/\sqrt{6} \\ 1/\sqrt{2} \amp -1/\sqrt{2} \amp 0 \\ 1/\sqrt{3} \amp 1/\sqrt{3} \amp 1/\sqrt{3} \end{amatrix} \text{.} \end{align*}
.

Example 5.2.20. Orthonormal change of basis: polynomials.

Consider the vector space \(P_1\) with inner product \(\langle p(x), q(x)\rangle=p(-1)q(-1)+p(1)q(1)\text{.}\) The ordered bases
\begin{align*} B\amp =\left(p_1(x)=\frac{1}{\sqrt{2}}x,p_2(x)=\frac{1}{\sqrt{2}}\right)\\ B'\amp=\left(q_1(x)=\frac{1}{2}(x-1), q_2(x)=\frac{1}{2}(x+1)\right) \end{align*}
are both orthonormal with respect to this inner product. Compute \(\underset{B\rightarrow B'}{P}\) and \(\underset{B'\rightarrow B}{P}\text{.}\)
Solution.
Since \(B'\) is orthonormal, we use Theorem 5.2.11 to compute
\begin{align*} [p_1(x)]_{B'} \amp = \left(\langle p_1(x),q_1(x)\rangle, \langle p_1(x),q_2(x)\rangle\right)\\ \amp =(p_1(-1)q_1(-1)+p_1(1)q_2(1),p_1(1)q_1(1)+p_1(1)q_2(1))=\frac{1}{\sqrt{2}}(1,1)\\ [p_2(x)]_{B'} \amp = \left(\langle p_2(x),q_1(x)\rangle, \langle p_1(x),q_2(x)\rangle\right)\\ \amp =(p_2(-1)q_1(-1)+p_2(1)q_2(1),p_2(1)q_1(1)+p_2(1)q_2(1))=\frac{1}{\sqrt{2}}(-1,1)\text{.} \end{align*}
Thus
\begin{equation*} \underset{B\rightarrow B'}{P}=\frac{1}{\sqrt{2}}\begin{amatrix}[rr] 1\amp -1\\ 1\amp 1 \end{amatrix} \end{equation*}
and by Theorem 5.2.18
\begin{equation*} \underset{B'\rightarrow B}{P}=(\underset{B\rightarrow B'}{P})^T=\frac{1}{\sqrt{2}}\begin{amatrix}[rr] 1\amp 1\\ -1\amp 1 \end{amatrix}\text{.} \end{equation*}

Exercises 5.2.4 Exercises

WeBWork Exercises

1.
Let \({ \mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 }\) be an orthonormal basis for an inner product space \(V\text{.}\) If
\begin{equation*} \mathbf{v} = a \mathbf{u}_1 + b \mathbf{u}_2 + c \mathbf{u}_3 \end{equation*}
is so that \(\|\mathbf{v}\| = 76\text{,}\) \(\mathbf{v}\) is orthogonal to \(\mathbf{u}_3\text{,}\) and \(\langle \mathbf{v},\mathbf{u}_2 \rangle = -76\text{,}\) find the possible values for \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\)
\(a =\) , \(b =\) , \(c =\)
Answer 1.
\(0\)
Answer 2.
\(-76\)
Answer 3.
\(0\)
Solution.
Solution: One checks by direct computation that
\(a = 0\text{,}\) \(b = -76\text{,}\) \(c = 0\)
must hold.
2.
Let
\begin{equation*} \vec{x} = \left[\begin{array}{c} 2\cr 3\cr -3 \end{array}\right] \ \mbox{ and } \ \vec{y} = \left[\begin{array}{c} 1\cr -5\cr 3 \end{array}\right]. \end{equation*}
Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of \({\mathbb R}^3\) spanned by \(\vec{x}\) and \(\vec{y}\text{.}\)
\(\Bigg\lbrace\) (3 × 1 array), (3 × 1 array) \(\Bigg\rbrace.\)
3.
Let
\begin{equation*} \vec{x} = \left[\begin{array}{c} 3\cr -3\cr 1\cr 0 \end{array}\right], \ \vec{y} = \left[\begin{array}{c} -3.5\cr 4.5\cr -4.5\cr 1 \end{array}\right], \ \vec{z} = \left[\begin{array}{c} 13\cr 8\cr 4\cr 10 \end{array}\right]. \end{equation*}
Use the Gram-Schmidt process to determine an orthonormal basis for the subspace of \({\mathbb R}^4\) spanned by \(\vec{x}\text{,}\) \(\vec{y}\text{,}\) and \(\vec{z}\text{.}\)
\(\Bigg\lbrace\) (4 × 1 array), (4 × 1 array), (4 × 1 array) \(\Bigg\rbrace.\)
4.
Let
\begin{equation*} A = \left[\begin{array}{cc} 3 \amp -9\cr -2 \amp 4\cr 1 \amp -7 \end{array}\right]. \end{equation*}
Find an orthonormal basis of the image of \(A\text{.}\)
\(\Bigg\lbrace\) (3 × 1 array), (3 × 1 array) \(\Bigg\rbrace.\)
5.
Let
\begin{equation*} A = \left[\begin{array}{cccc} -12 \amp 9 \amp -8 \amp 6\cr -3 \amp -6 \amp -2 \amp -4 \end{array}\right]. \end{equation*}
Find an orthonormal basis of the kernel of \(A\text{.}\)
\(\Bigg\lbrace\) (4 × 1 array), (4 × 1 array) \(\Bigg\rbrace.\)

6.

The vectors
\begin{equation*} \boldv_1=(1,1,1,1), \boldv_2=(1,-1,1,-1), \boldv_3=(1,1,-1,-1), \boldv_4=(1,-1,-1,1) \end{equation*}
are pairwise orthogonal with respect to the dot product, as is easily verified. For each \(\boldv\) below, find the scalars \(c_i\) such that
\begin{equation*} \boldv=c_1\boldv_1+c_2\boldv_2+c_3\boldv_3+c_4\boldv_4\text{.} \end{equation*}
  1. \(\displaystyle \boldv=(3,0,-1,0)\)
  2. \(\displaystyle \boldv=(1,2,0,1)\)
  3. \(\boldv=(a,b,c,d)\) (Your answer will be expressed in terms of \(a,b,c\text{,}\) and \(d\text{.}\) )

Coordinate vectors: orthogonal basis.

In each exercise an inner product space \((V,\langle\, \rangle)\) and orthogonal ordered basis is given. Use Theorem 5.2.11 to compute the requested coordinate vector.
7.
\(V=\R^3\) with dot product; \(B=\left((1,1,1),(1,-1,0),(1,1,-2)\right)\text{.}\) Compute \([(-3,2,4)]_B\text{.}\)
8.
\(V=\R^3\) with dot product with weights \(k_1=1, k_2=2, k_3=2\text{;}\) \(B=\left((1,1,1),(2,1,-2),(4,-3,1)\right)\text{.}\) Compute \([(0,1,0)]_B\text{.}\)
9.
\(V=\Span\{\cos x, \cos 2x, \cos 3x\}\subseteq C([0,2\pi])\) with integral inner product \(\langle f, g\rangle=\int_0^{2\pi}f(x)g(x)\, dx\text{;}\) \(B=(\cos x, \cos 2x, \cos 3x)\text{.}\) Compute \([\cos^3 x]_B\text{.}\) (Yes, \(\cos^3(x)\) can indeed be written as a linear combination of \(\cos x, \cos 2x, \cos 3x\text{.}\) In this exercise you will discover what the corresponding identity is using inner products!)

Orthonormal change of basis.

In each exercise an inner product space \((V,\langle\, , \rangle)\) is given along with two orthonormal ordered bases \(B\) and \(B'\text{.}\) Compute \(\underset{B\rightarrow B'}{P}\) and \(\underset{B'\rightarrow B}{P}\) using Theorem 5.2.18.
10.
\(V=\R^2\) with the dot product, \(B=\left((\sqrt{2}/2, \sqrt{2}/2), (-\sqrt{2}/2, \sqrt{2}/2)\right)\text{,}\) \(B'=\left( (\sqrt{3}/2, -1/2), (1/2,\sqrt{3}/2)\right)\)
11.
\(V=\R^4\) with the dot product, \(B=(\bolde_1, \bolde_2, \bolde_3, \bolde_4)\text{,}\) \(B'=\left(\frac{1}{2}(1,1,1,1), \frac{1}{2}(1,-1,1,-1), \frac{1}{2}(1,1,-1,-1), \frac{1}{2}(1,-1,-1,1)\right)\)

Gram-Schmidt procedure.

Use the Gram-Schmidt procedure to convert the given basis to a basis that is orthogonal with respect to the given inner product.
12.
\(V=\R^3\) with the weighted dot product
\begin{equation*} \langle (x_1,x_2,x_3),(y_1,y_2,y_3)\rangle=x_1y_1+2x_2y_2+3x_3y_3\text{.} \end{equation*}
\(B=\{(1,1,0),(0,1,1), (1,0,1)\}\text{.}\)
13.
\(V=\Span \{x^2,x^4,x^6\}\subseteq C([0,1])\) with the integral inner product
\begin{equation*} \langle f, g\rangle=\int_0^1 f(x)g(x)\, dx\text{.} \end{equation*}
\(B=\{x^2,x^4,x^6\}\text{.}\)
14.
\(V=P_2\) with the evaluation inner product
\begin{equation*} \angvec{p(x),q(x)}=p(-1)q(-1)+p(0)q(0)+p(1)q(1)\text{.} \end{equation*}
\(B=\{x^2, x, 1\}\text{.}\)

15.

Consider the inner product space \(\R^4\) together with the dot product.
\begin{equation*} W=\{\boldx\in \R^4\colon \boldx\cdot (1,2,-1,-1)=0\}\text{.} \end{equation*}
  1. Show that \(W\) is a subspace of \(\R^4\) by finding a matrix \(A\) for which \(W=\NS A\text{.}\)
  2. Use (a) and an appropriate fundamental space algorithm to find a basis for \(W\text{.}\)
  3. Use Gram-Schmidt to convert your basis in (b) to an orthgonal basis of \(W\text{.}\)

16. Extending orthogonal bases.

Consider the inner product space given by \(\R^4\) together with the dot product. Construct an orthogonal basis of \(\R^4\) containing \(\boldv_1=(1,1,1,1)\) following the steps below.
  1. Produce a vector \(\boldv_2\) orthogonal to \(\boldv_1\) by inspection.
  2. Produce a vector \(\boldv_3\) orthogonal to \(\boldv_1\) and \(\boldv_2\) by setting up an appropriate matrix equation of the form \(A\boldx=\boldzero\) and finding a nontrivial solution. (Use Theorem 5.1.9.)
  3. Produce a vector \(\boldv_4\) orthogonal to \(\boldv_1, \boldv_2, \boldv_3\) by setting up an appropriate matrix equation of the form \(B\boldx=\boldzero\) and finding a nontrivial solution. (Use Theorem 5.1.9.)

17. Extending orthogonal bases.

Consider the inner product space given by \(V=\R^3\) together with the dot product. Let \(W\) be the plane with defining equation \(x+2y-z=0\text{.}\) Compute an orthogonal basis of \(W\text{,}\) and then extend this to an orthogonal basis of \(\R^3\text{.}\)
Hint.
You do not have to use Gram-Schmidt here, but can proceed using a combination of inspection, your geometric understanding of \(W\text{,}\) and/or along similar lines of Exercise 5.2.4.16.

18.

Let \((V,\langle , \rangle )\) be an inner produce space. Prove: if \(\angvec{\boldv,\ \boldw}=0\text{,}\) then
\begin{equation*} \norm{\boldv+\boldw}^2=\norm{\boldv}^2+\norm{\boldw}^2\text{.} \end{equation*}
This result can be thought of as the Pythagorean theorem for general inner product spaces.

19.

Let \((V, \langle , \rangle )\) be an inner product space, and suppose \(B=\{\boldv_1, \boldv_2, \dots, \boldv_n\}\) is an orthonormal basis of \(V\text{.}\) Suppose \(\boldv, \boldw\in V\) satisfy
\begin{equation*} \boldv=\sum_{i=1}^nc_i\boldv_i, \boldw=\sum_{i=1}^nd_i\boldv_i\text{.} \end{equation*}
  1. Prove:
    \begin{equation*} \langle \boldv, \boldw\rangle =\sum_{i=1}^nc_id_i\text{.} \end{equation*}
  2. Prove:
    \begin{equation*} \norm{\boldv}=\sqrt{\sum_{i=1}^nc_i^2}\text{.} \end{equation*}

20. Orthonormal coordinate vectors.

Let \((V, \langle\, , \rangle)\) be an inner product space, and suppose \(B=(\boldv_1, \boldv_2, \dots, \boldv_n)\) is an orthonormal ordered basis of \(V\text{.}\)
  1. Prove that
    \begin{equation*} \langle \boldv, \boldw\rangle =[\boldv]_B\cdot [\boldw]_B \end{equation*}
    for all \(\boldv, \boldw\in V\text{.}\) In other words we can compute the inner product of vectors by computing the dot product of their coordinate vectors with respect to the orthonormal basis \(B\text{.}\)
  2. Prove that a set \(S=\{\boldw_1, \boldw_2, \dots, \boldw_r\}\subseteq V\) is orthogonal (resp. orthonormal) with respect to \(\langle\, , \rangle\) if and only if \(S'=\{[\boldw_1]_B, [\boldw_2]_B, \dots, [\boldw_r]_B\subseteq \R^n\) is orthogonal (resp. orthonormal) with respect to the dot product.

21. Determinant of orthogonal matrices.

Prove: if \(Q\in M_{nn}\) is an orthogonal matrix, then \(\det Q=\pm 1\text{.}\)

22. Orthogonal \(2\times 2\) matrices.

In this exercise we will prove that a \(2\times 2\) matrix is orthogonal if and only it is a rotation matrix or a reflection matrix.
Let \(Q\) be a \(2\times 2\) matrix.
  1. Prove that if \(Q\) is orthogonal and \(\det Q=1\text{,}\) then \(Q\) is a rotation matrix: i.e., there is a \(\theta\in [0,2\pi]\) such that
    \begin{equation*} Q=\begin{amatrix}[rr] \cos\theta\amp -\sin\theta\\ \sin\theta\amp \cos\theta \end{amatrix}\text{.} \end{equation*}
    See Theorem 3.2.13.
  2. Prove that if \(Q\) is orthogonal and \(\det Q=-1\text{,}\) then \(Q\) is a reflection matrix: i.e., there is a \(\theta\in [0,\pi]\) such that
    \begin{equation*} Q=\begin{amatrix}[rr] \cos 2\theta\amp \sin 2\theta\\ \sin 2\theta \amp -\cos 2\theta \end{amatrix}\text{.} \end{equation*}
    See Theorem 3.2.17.
  3. Prove that \(Q\) is an orthogonal matrix if and only if \(Q\) is a rotation matrix or \(Q\) is a reflection matrix. You may use the result of Exercise 5.2.4.21.
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