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Power of matrix

Let \displaystyle A=\begin{pmatrix} -2 & 4 &3 \\ 0 & 0 & 0\\ -1 &5 &2 \end{pmatrix}. Prove that A^{593}-2A^{15}+A=0.

Solution

The characteristic polynomial of A is p(x)=x-x^3. This in return means A=A^3 and A^3=A^5. Thus,

    \begin{align*} A^{593} -2 A^{15} +A &= A^{591} \cdot A^2 - 2 \left ( A^3 \right )^5 + A\\ &=\left ( A^3 \right )^{197} \cdot A^2 - 2 A^5 + A \\ &= A^{197} \cdot A^2 - 2 A^3 + A \\ &=A^{199} - 2 A +A\\ &=A^{198} \cdot A- A\\ &=\left ( A^3 \right )^{66} \cdot A - A \\ &=A^{66} \cdot A - A\\ &= \left ( A^3 \right )^{22} \cdot A - A\\ &= A^{22} \cdot A -A \\ &= A^{23} - A\\ &= \left ( A^3 \right )^{7} \cdot A^2 -A\\ &= A^7 \cdot A^2 - A\\ &= A^9 - A\\ &=\left ( A^3 \right )^3 - A\\ &=A^3 -A\\ &=A -A\\ &=\mathbb{O} \end{align*}

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Linear Projection

Let \mathcal{V} be a linear space over \mathbb{R} such that \dim_{\mathbb{R}} \mathcal{V} < \infty and f:\mathcal{V} \rightarrow \mathcal{V} be a linear projection such that any non zero vector of \mathcal{V} is an eigenvector of f. Prove that there exists \lambda \in \mathbb{R} such that f = \lambda \; \mathrm{Id} where \mathrm{Id} is the identity endomorphism.

Solution

Let \dim_{\mathbb{R}} \mathcal{V} = n. Since any non zero vector is an eigenvector it follows that every basis of \mathcal{V} is also an eigenbasis. Let {\cal{B}}=\bigl\{{\overrightarrow{e_{i}}}\bigr\}_{i=1}^{n} be such a basis and \lambda_i \;, \; i=1, 2, \dots, n be the respective , not necessarily distinct , eigenvalues of the eigenvectors of \mathcal{B}. For the vector \overrightarrow{y}=\mathop{\sum}\limits_{i=1}^{n}\overrightarrow{e_{i}} which also happens to be eigenvector with eigenvalue \lambda , it holds that:

    \begin{align*} f\left ( \vec{y} \right ) = \lambda \vec{y} &\Rightarrow f\left ( \sum_{i=1}^{n} \overrightarrow{e_i} \right ) = \lambda \sum_{i=1}^{n} \overrightarrow{e_i} \\ &\!\!\!\!\!\!\!\!\overset{f \; \text{linear}}{=\! =\! =\! =\! =\!\Rightarrow } \sum_{i=1}^{n} f \left ( \overrightarrow{e_i} \right ) = \lambda \sum_{i=1}^{n} \overrightarrow{e_i} \\ &\Rightarrow \sum_{i=1}^{n} \lambda_i \overrightarrow{e_i} = \lambda \sum_{i=1}^{n} \overrightarrow{e_i} \\ &\Rightarrow \sum_{i=1}^{n} \left ( \lambda_i - \lambda \right ) \overrightarrow{e_i} =0 \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{e_i \;\; \text{linearly independent}}{=\! =\! =\! =\! =\! =\!=\! =\!=\!=\!=\!=\!=\!\Rightarrow } \left [ \left ( \forall i=1, 2, \dots, n \right ) \;\; \lambda_i - \lambda =0 \right ] \\ &\Rightarrow \left [ \left ( \forall i=1, 2, \dots, n \right ) \;\; \lambda_i = \lambda \right ] \end{align*}

But then for each \overrightarrow{x}=\mathop{\sum}\limits_{i=1}^{n}x_i\overrightarrow{e_{i}}\in{\cal{V}} it holds that

    \[f\bigl({\overrightarrow{x}}\bigr)=f\left({\sum_{i=1}^{n}x_i\overrightarrow{e_{i}}}\right)=\displaystyle\sum_{i=1}^{n}x_if\bigl({\overrightarrow{e_{i}}}\bigr)=\sum_{i=1}^{n}x_i\lambda\,\overrightarrow{e_{i}}=\lambda\sum_{i=1}^{n}x_i\overrightarrow{e_{i}}=\lambda\overrightarrow{x}\]

The result follows.

Remark: This proof also works in the case \mathcal{V} is infinite.

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Convexity of exponential function

If C, D are n \times n symmetric real matrices we write C \geq D if-f the matrix C-D is non negative definite. Examine if

    \[\exp \left ( \frac{A+B}{2} \right ) \leq \frac{\exp A +\exp B}{2}\]

for each pair 2 \times 2 real symmetric matrices A, B such that AB=BA.

Solution

Since A, B commute it follows from the properties of the exponential function

(1)   \begin{equation*} \exp{\frac{A+B}{2}}=\exp{\frac{A}{2}} \exp{\frac{B}{2}} \end{equation*}

Noting that

(2)   \begin{equation*} \frac{\exp A +\exp B}{2}-\exp \left ( \frac{A+B}{2} \right )=\frac{1}{2}\left(\exp {\frac{A}{2}}-\exp {\frac{B}{2}} \right)^{2}  \end{equation*}

Setting \displaystyle M= \exp {\frac{A}{2}}-\exp {\frac{B}{2}} we must prove that M^2 \geq 0. Since A, B commute so are \displaystyle \exp \frac{A}{2} \; , \; \exp \frac{B}{2}. Hence M is symmetric. Thus,

(3)   \begin{equation*} x^{\top }M^{2}x=x^{\top }M^{\top }Mx=\left \| Mx \right \|^{2}  \end{equation*}

and that’s all!

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Zero matrix

Let A\in \mathcal{M}_n\left ( \mathbb{C} \right ) with n\geq 2. If \det \left ( A+X \right )=\det A+\det X for every matrix X \in \mathcal{M}_n\left ( \mathbb{C} \right ) then prove that A=\mathbb{O}_{n}.

Solution

Suppose that A \neq 0, say A_{ij} \neq 0 for some i,j. Let P be any permutation matrix with P_{ij}=1 and let Q be the matrix obtained from P by changing its ij-entry to 0. Finally let X = xQ where x \in \mathbb{C}.

We have that \det X = 0 and that \det X = \det(A+X) - \det A is a polynomial in x. Furthermore, the coefficient of x^{n-1} of this polynomial is \pm A_{ij} depending on the sign of the corresponding permutation. So the polynomial is not identically zero, a contradiction.

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Invertible matrix

Consider the matrices A \in \mathcal{M}_{m \times n} and B \in \mathcal{M}_{n \times m}. If AB +\mathbb{I}_m is invertible prove that BA+\mathbb{I}_n is also invertible.

Solution

So we have to answer the question if -1 is a zero of the essentially same characteristic polynomials. AB and BA have quite similar characteristic polynomials. In fact if p(x) denotes the polynomial of AB, then the polynomial of BA will be q(x)= x^{n-m} p(x). It is easy to see that -1 cannot be an eigenvalue of the AB matrix, otherwise it wouldn’t be invertible. Now, let us assume that BA is not invertible. Then it must have an eigenvalue of -1 and let \mathbf{x} be the corresponding eigenvector. Hence:

    \[\left ( BA \right )\mathbf{x}= -\mathbf{x} \Rightarrow AB \left ( A \mathbf{x} \right )= -A\mathbf{x}\]

meaning that AB has an eigenvalue of -1 which is a contradiction. The result follows.

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