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 $A^{-1}=\begin{bmatrix}1&-2&4\\0&1&-2\\0&0&1\end{bmatrix}$.

  • $\begingroup$ So is the question here to show eigenvalues of $A^tA$ belong to $(0,8)$? $\endgroup$ – Yadati Kiran Nov 14 '18 at 10:24
  • $\begingroup$ 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. $\endgroup$ – Adrien Laurent Nov 14 '18 at 10:26
  • $\begingroup$ @Yadati Kiran yes it is ! $\endgroup$ – James Well Nov 14 '18 at 12:15
  • $\begingroup$ 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 $\sum|\lambda-a_{ii}|\leq\sum R_i\implies |3\lambda-11|\leq8 \implies 1\leq\lambda\leq\dfrac{19}{3}.$ $\endgroup$ – 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 $\lambda < 8$.

  • $\begingroup$ 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 $||A^tA||$ ? $\endgroup$ – 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 $\lambda < 8$.

  • $\begingroup$ 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 $A^tA$ $\endgroup$ – James Well Nov 14 '18 at 23:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.