Problems

01

Finding a spectral decomposition

Let $A=\left(\begin{array}{ccc}4& 2& 2\\ 2& 4& 2\\ 2& 2& 4\end{array}\right)$.

• (a) Find the eigenvalues of $A$. There will be two distinct numbers, ${\lambda }_1,{\lambda }_2$, not three.
• (b) Row reduce $A_{{\lambda }_1}$ and $A_{{\lambda }_2}$ to solve the two equations $A_{{\lambda }_1}\mathbf{x}=0$ and $A_{{\lambda }_2}\mathbf{x}=0$. In this way, write down a set of 3 eigenvectors, two of which have one of the eigenvalues, and the third has the other eigenvalue.
• (c) For the ‘double’ eigenvalue, convert its two eigenvectors to orthogonal vectors (subtract a projection like in Gram-Schmidt, unless they are already perpendicular).
• (d) Normalize all your orthogonal eigenvectors to get unit eigenvectors. Call them ${\mathbf{q}}_1,{\mathbf{q}}_2,{\mathbf{q}}_3$. Now the matrix $Q=\left(\begin{array}{ccc}{\mathbf{q}}_1& {\mathbf{q}}_2& {\mathbf{q}}_3\end{array}\right)$ is an orthogonal matrix, its inverse is $Q^{-1}=Q^{\mathsf{T}}$, and we know that $A=QD{Q}^{\mathsf{T}}$ where $D$ has the eigenvectors ${\lambda }_1,{\lambda }_2,{\lambda }_3$ along the diagonal. (In order, so $A{\mathbf{q}}_i={\lambda }_i{\mathbf{q}}_i$ for $i=1,2,3$.)
• (e) Write out the spectral decomposition: $A={\lambda }_1{\mathbf{q}}_1{\mathbf{q}}_1^{\mathsf{T}}+{\lambda }_2{\mathbf{q}}_2{\mathbf{q}}_2^{\mathsf{T}}+{\lambda }_3{\mathbf{q}}_3{\mathbf{q}}_3^{\mathsf{T}}$. In other words, simply compute the three projection matrices ${\mathbf{q}}_i{\mathbf{q}}_i^{\mathsf{T}}$, and plug them into this formula. Check that if you were to take the sum, you would get $A$.

02

Quadratic form

Let $A$ be the matrix as in Problem 12-01. Use results of 12-01 for this problem: you do not need to find the eigenvalues / eigenvectors again.

• (a) What is the quadratic form $Q(\mathbf{x})$ associated to $A$?
• (b) What is the quadratic form $Q(\mathbf{y})$ in the variables $\mathbf{y}$ that eliminate cross terms?
• (c) What is the max and min of $Q(\mathbf{x})$ for $|\mathbf{x}|=1$?
• (d) What are the vectors $\mathbf{x}$ that yield the max and min found in (d)?

03

Functions of symmetric matrices

• (a) Write out in coefficients the matrix $A=7{\mathbf{q}}_1{\mathbf{q}}_1^{\mathsf{T}}+{\mathbf{q}}_2{\mathbf{q}}_2^{\mathsf{T}}+3{\mathbf{q}}_3{\mathbf{q}}_3^{\mathsf{T}}$ when ${\mathbf{q}}_i={\mathbf{e}}_i$.
• (b) Compute the matrix product $A^2$ when ${\mathbf{q}}_i={\mathbf{e}}_i$. (Notice that you have found $B=A^2$.)
• (c) Write out in coefficients the matrix $A=49{\mathbf{q}}_1{\mathbf{q}}_1^{\mathsf{T}}+{\mathbf{q}}_2{\mathbf{q}}_2^{\mathsf{T}}+9{\mathbf{q}}_3{\mathbf{q}}_3^{\mathsf{T}}$ when ${\mathbf{q}}_i={\mathbf{e}}_i$.
• (d) Guess the coefficients ${\lambda }_1,{\lambda }_2,{\lambda }_3$ in $B={\lambda }_1{\mathbf{q}}_1{\mathbf{q}}_1^{\mathsf{T}}+{\lambda }_2{\mathbf{q}}_2{\mathbf{q}}_2^{\mathsf{T}}+{\lambda }_3{\mathbf{q}}_3{\mathbf{q}}_3^{\mathsf{T}}$ such that $B^2=A$ for $A$ as in (a). Verify your guess by writing out $B$ in coefficients and then calculating $B^2$. (Notice that you have found $B=\sqrt{A}$.)
• (e) Now drop the assumption that ${\mathbf{q}}_i={\mathbf{e}}_i$, and yet find $A^2$ by FOILing the product $(7{\mathbf{q}}_1{\mathbf{q}}_1^{\mathsf{T}}+{\mathbf{q}}_2{\mathbf{q}}_2^{\mathsf{T}}+3{\mathbf{q}}_3{\mathbf{q}}_3^{\mathsf{T}})(7{\mathbf{q}}_1{\mathbf{q}}_1^{\mathsf{T}}+{\mathbf{q}}_2{\mathbf{q}}_2^{\mathsf{T}}+3{\mathbf{q}}_3{\mathbf{q}}_3^{\mathsf{T}})$. To simplify the result, figure out what ${\mathbf{q}}_i{\mathbf{q}}_i^{\mathsf{T}}{\mathbf{q}}_j{\mathbf{q}}_j^{\mathsf{T}}$ is for $i=j$ and $i\ne j$.
• (f) Based on (a)-(e), can you guess what ${\lambda }_1,{\lambda }_2,{\lambda }_3$ should be in the matrix $e^A$? (Hint: consider that $e^x$ has a power series expansion: $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$.)

04

Positive definite matrices

Suppose $A^{\mathsf{T}}=A$ and $A^{-1}$ exists. (Symmetric and invertible.) Notice that the Spectral Theorem applies.

• (a) What are the eigenvalues of $A^{-1}$ in terms of the eigenvalues of $A$?
• (b) Suppose that $Q(\mathbf{x})={\mathbf{x}}^{\mathsf{T}}A\mathbf{x}$ is positive definite. Show that $Q^{\prime }(\mathbf{x})={\mathbf{x}}^{\mathsf{T}}{A}^{-1}\mathbf{x}$ is also positive definite. (Hint: make use of (a).)