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

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  • $\begingroup$ 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
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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 $\lambda < 8$.

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  • $\begingroup$ 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
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$A^tA$ 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 $\lambda < 8$.

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  • $\begingroup$ 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

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