Let $V=\mathbb{R}^3$ and $u_1=(1,0,1), \quad u_2=(2,3,2), \quad u_3=(-1,4,7).$
Compute a basis $v_1,v_2,v_3$ for $\mathbb{R}^3$ which is orthonormal with respect to the dot product such that $\operatorname{Sp}\left\{u_{1}, \cdots, u_{i}\right\}=\operatorname{Sp}\left\{v_{1}, \cdots, v_{i}\right\}$ for each $1 \leq i \leq 3$. Solution.(The Gram-Schmidt procedure)
$v_1=\frac{u_1}{\|u_1\|}=\left(\frac1{\sqrt2},0,\frac1{\sqrt2}\right),w_2=u_2-(u_2·v_1)v_1=(0,3,0),v_2=\frac{w_2}{\|w_2\|}=(0,1,0),$$w_3=u_3-(u_3·v_1)v_1-(u_3·v_2)v_2=(-4,0,4),v_3=\frac{w_3}{\|w_3\|}=\left(-\frac1{\sqrt2},0,\frac1{\sqrt2}\right)$
Let $A \in M_{n}(\mathbb{R})$ be a symmetric matrix. Show that if $\lambda, \mu \in \mathbb{R}$ are distinct eigenvalues of $A$ with $v$ and $w$ associated eigenvectors, then $v$ and $w$ are orthogonal; that is, $v^{\sf T} w=0$. Proof.
From $Av=λv⇒v^{\sf T}A=λv^{\sf T}⇒v^{\sf T}Aw=λv^{\sf T}w$ and $Aw=μw⇒v^{\sf T}Aw=μv^{\sf T}w$, we have $(λ-μ)v^{\sf T}w=0$, but $λ-μ≠0$, so $v^{\sf T}w=0$.
Find a real orthogonal matrix $P$ such that $P^{\sf T} A P$ is diagonal when $A$ is each of the following matrices
$\left(\begin{array}l
3 & 2 \\
2 & 0
\end{array}\right),\left(\begin{array}l
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right),\left(\begin{array}l
1 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right)$ Solution.
(1)Eigenvalues:$-1,4$. Eigenvectors:$\frac1{\sqrt5}\begin{pmatrix}1\\-2\end{pmatrix},\frac1{\sqrt5}\begin{pmatrix}2\\1\end{pmatrix}$. $P=\begin{pmatrix}\frac1{\sqrt5}&\frac2{\sqrt5}\\-\frac2{\sqrt5}&\frac1{\sqrt5}\end{pmatrix}$
(2)Eigenvalues:$0,3$. Eigenvectors:$\frac1{\sqrt2}\begin{pmatrix}1\\-1\\0\end{pmatrix},\frac1{\sqrt6}\begin{pmatrix}1\\1\\-2\end{pmatrix},\frac1{\sqrt3}\begin{pmatrix}1\\1\\1\end{pmatrix}$. $P=\begin{pmatrix}\frac1{\sqrt2}&\frac1{\sqrt6}&\frac1{\sqrt3}\\-\frac1{\sqrt2}&\frac1{\sqrt6}&\frac1{\sqrt3}\\0&-\frac2{\sqrt6}&\frac1{\sqrt3}\end{pmatrix}$.
(3)Eigenvalues:$1,\frac{1+\sqrt5}2,\frac{1-\sqrt5}2$. Eigenvectors:$\begin{pmatrix}0\\1\\0\end{pmatrix},\begin{pmatrix}\frac{1+\sqrt5}2\sqrt{5-\sqrt5\over10}\\0\\\sqrt{5-\sqrt5\over10}\end{pmatrix},\begin{pmatrix}\frac{1-\sqrt5}2\sqrt{5+\sqrt5\over10}\\0\\\sqrt{5+\sqrt5\over10}\end{pmatrix}$. $P=\begin{pmatrix}0&\frac{1+\sqrt5}2\sqrt{5-\sqrt5\over10}&\frac{1-\sqrt5}2\sqrt{5+\sqrt5\over10}\\1&0&0\\0&\sqrt{5-\sqrt5\over10}&\sqrt{5+\sqrt5\over10}\end{pmatrix}$.
Remark: By Q2, it suffices to pick orthonormal basis for each individual eigenspace.
Verify that if $P$ is an orthogonal matrix and $x=Py$ then $y^{\sf T} y=x^{\sf T} x$.
Let $A$ be a real symmetric $n \times n$ matrix. Then we know that there exists a real orthogonal matrix $P$ such that $P^{\sf T} A P$ is diagonal. By using the transformation $x=P y$, or otherwise, prove that for every $x \in \mathbb{R}^{n}$, $m x^{\sf T} x \leq x^{\sf T} A x \leq M x^{\sf T} x,$ where $m$ and $M$ are the smallest and greatest eigenvalues of $A$ respectively. For which $x$ is it true that $x^{\sf T} A x=M x^{\sf T} x$ ?
Let $A=\left(\begin{array}r5 & 1 & \sqrt{2} \\ 1 & 5 & \sqrt{2} \\ \sqrt{2} & \sqrt{2} & 6\end{array}\right)$. Find the maximum and minimum values of $x^{\sf T} x$ for those $x$ for which $x^{\sf T} A x=1$. Giving no heed to orientation, sketch the surface $S$ with equation $x^{\sf T} A x=1$, and indicate on it those vectors $x$ at which $x^{\sf T} x$ attains its maximum and minimum values on $S$. Proof.
$x=Py$⇒$x^{\sf T}=y^{\sf T}P^{\sf T}$⇒$x^{\sf T}x=y^{\sf T}P^{\sf T}Py=y^{\sf T}Iy=y^{\sf T}y$.
Let $P^{\sf T} A P=\diag(λ_1,λ_2,⋯,λ_n),m\le λ_1,λ_2,⋯λ_n\le M$, then $x^{\sf T}Ax=(Py)^{\sf T}A(Py)=y^{\sf T}(P^{\sf T}AP)y=λ_1y_1^2+⋯+λ_ny_n^2$ and $mx^{\sf T}x=my^{\sf T}y=m(y_1^2+⋯+y_n^2)≤λ_1y_1^2+⋯+λ_ny_n^2≤M(y_1^2+⋯+y_n^2)=My^{\sf T}y=Mx^{\sf T}x$. Equality: $x$ is eigenvector for $m,M$.
Let $M=λ_k,1≤k≤n$, for $x$ collinear with $P\mathbf e_k$, it is true that $x^{\sf T} A x=M x^{\sf T} x$.
The eigenvalues of $A$ are 4,8. The maximum of $x^{\sf T}x$ for those $x$ for which $x^{\sf T}Ax$=1 is $\frac14$ and minimum is $\frac18$. The surface is an ellipsoid (specifically, an oblate spheroid) with semiaxis $1\over2$,$1\over2$ and $1\over2\sqrt{2}$.
Show that for any real $n\times n$ matrix $A,A^{\sf T}A$ is symmetric.
Suppose that $A=\begin{pmatrix}
1 & 0 & 1 \\
1 & 1 & 0 \\
0 & 1 & -1
\end{pmatrix}.$ By considering $A^{\sf T} A$, or otherwise, calculate the maximum and minimum value of $\|A\mathbf x\|$ on the unit sphere. Proof.
$(A^{\sf T}A)^{\sf T}=A^{\sf T}(A^{\sf T})^{\sf T}=A^{\sf T}A⇒A^{\sf T}A$ is symmetric.
The eigenvalues of $A^{\sf T}A=\left(\begin{array}c2 & 1 & 1 \\ 1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right)$ are 3,3,0 and corresponding eigenvectors are (1,0,1),(1,1,0),(-1,1,1), by Q4, the maximum value of $\|A\mathbf x\|$ on the unit sphere is $\sqrt3$(when $\mathbf x=\cosθ\left(\frac1{\sqrt2},0,\frac1{\sqrt2}\right)+\sinθ\left(\frac1{\sqrt2},\frac1{\sqrt2},0\right)$) and minimum is 0(when $\mathbf x=±\left(-\frac1{\sqrt3},\frac1{\sqrt3},\frac1{\sqrt3}\right)$).
Remark: $\|A\mathbf x\|=1$ is a circular cylinder (since we can find a rotation $(x,y)\mapsto(x',y')$ such that $3x'^2+3y'^2=\|A\mathbf x\|^2$). When the cylinder degenerates into a straight line through the origin, $\|A\mathbf x\|$ reaches minimum 0. When the cylinder base radius increases to 1, it is tangent to the sphere and the intersection path is a big circle on the sphere and $\|A\mathbf x\|$ reaches maximum $\sqrt3$.