# Prove eigenvalues of a symmetric matrix are in a certain interval

I am given a matrix $$A=\begin{bmatrix}1&2&0\\0&1&2\\0&0&1\end{bmatrix}$$. I am asked to compute $$A^tA=\begin{bmatrix}1&2&0\\2&5&2\\0&2&5\end{bmatrix}$$

and then to prove that the eigenvalues of $$A^tA$$ are all in $$]0,8[$$. I have no idea how to prove this. I think it has something to do with matrix norms ?

I don't know whether the previous questions are relevant here, but I was asked to compute

$$e^{tA}=\begin{bmatrix}e^t&2te^t&2t^2e^t\\0&e^t&2te^t\\0&0&e^t\end{bmatrix}$$ and $$A^{-1}=\begin{bmatrix}1&-2&4\\0&1&-2\\0&0&1\end{bmatrix}$$.

• So is the question here to show eigenvalues of $A^tA$ belong to $(0,8)$? – Yadati Kiran Nov 14 '18 at 10:24
• Using Gershgorin's theorem (en.wikipedia.org/wiki/Gershgorin_circle_theorem) yields that the eigenvalues are in [-1,9]. It's almost what you want. ;) Also you can compute the eigenvalues directly using the characteristic polynomial. – Adrien Laurent Nov 14 '18 at 10:26
• @Yadati Kiran yes it is ! – James Well Nov 14 '18 at 12:15
• We can do something more with Gershgorin's theorem. We get the following $|\lambda-1|\leq2, |\lambda-|\leq4, |\lambda-5|\leq2$. We can also say $\sum|\lambda-a_{ii}|\leq\sum R_i\implies |3\lambda-11|\leq8 \implies 1\leq\lambda\leq\dfrac{19}{3}.$ – Yadati Kiran Nov 14 '18 at 12:47

If $$A^TAx = \lambda x$$, then $$x^TA^TAx = \lambda x^Tx$$, so $$\lambda ||x||^2 = ||Ax||^2$$, where $$||y|| = \sqrt{\sum y_i^2}$$ is the norm of the vector $$y$$. It follows that $$\lambda > 0$$, since $$A$$ is invertible, so $$||Ax|| \neq 0$$ for $$x \neq 0$$.

To see that $$\lambda < 8$$, we need to show that for all $$x$$, we have $$||Ax||^2 < 8 ||x||^2$$. For this, note that $$x = (a,b,c) \implies Ax = (a+2b,b+2c,c)$$, in which case $$||Ax||^2 = a^2 + 4b^2 + 4ab + b^2 + 4c^2 + 4bc + c^2 = a^2 + 5b^2 + 5c^2 + 4ab+4bc$$

Take the difference $$8||x||^2 - ||Ax||^2$$, it is equal to $$7a^2+3b^2+3c^2 -4ab-4bc$$. Can we prove this is greater than zero for all $$(a,b,c)$$ non-zero?

Well, we can, by combining the $$ab$$ and $$bc$$ nicely into squares.Like this: $$(2b^2 -4bc+2c^2) + (4a^2 - 4ab+b^2) + c^2 + 3a^2 = 2(b-c)^2 + (2a-b)^2 + c^2 + 3a^2$$

which is a positive linear combination of squares. Note that if the RHS equals zero, this forces $$c=a=0$$ and $$b-c = 0$$ so $$b = 0$$. In other words, if $$(a,b,c) \neq (0,0,0)$$ then the difference is positive, giving $$\lambda < 8$$.

• I do like this solution. There's one thing I don't get, why are we proving that $||Ax||^2<8||x||^2$ and not $||A^tAx||<8||x||$ in order to bound the eigenvalues of $||A^tA||$ ? – James Well Nov 15 '18 at 0:23

$$A^tA$$ is a positive definite symmetric matrix. Therefore, its singular values coincide with its eigenvalues [1]. This gives $$\lambda > 0$$.

According to Wikipedia, we have $$\|A\|_{2}=\sigma _{\max }(A)\leq \left(\sum _{i=1}^{m}\sum _{j=1}^{n}|a_{ij}|^{2}\right)^{1/2}=\|A\|_{\rm {F}}$$ The equality holds if and only if $$A$$ is a rank-one matrix or the zero matrix.

For the matrix in question, we have $$\|A\|_{\rm {F}}=8$$. This gives $$\lambda < 8$$.

• Tell me if I'm wrong, but it seems to me like you've proven that the eigenvalues of $A$ are in $]0,8[$, not those of $A^tA$ – James Well Nov 14 '18 at 23:05