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Tag Archives: Linear Algebra

Determinant

Let x>0 and  A \in \mathbb{R}^{2 \times 2} such that \det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = 0. Prove that

    \[\det \left( A^2 + A + x \mathbb{I}_{2 \times 2} \right) = x\]

Solution

Note that

    \[\det \left(A^2 + x \mathbb{I}_{2 \times 2} \right) = \det \left(A + i \sqrt {x} \mathbb{I}_{2 \times 2} \right) \det \left(A - i\sqrt{x}  \mathbb{I}_{2 \times 2} \right)\]

Since A is real, its complex eigenvalues come in conjugate pairs. Thus, in this case we conclude that A has eigenvalues \pm i \sqrt x.

Now, if \lambda is an eigenvalue of A, then \lambda^2 + \lambda + x is an eigenvalue of A^2 + A + x \mathbb{I}_{2 \times 2}. Thus, the matrix A^2 + A + x \mathbb{I}_{2 \times 2} has eigenvalues (i\sqrt {x} )^2 + i\sqrt {x} + x = i\sqrt {x} and (-i\sqrt {x})^2 - i\sqrt {x} + x = -i\sqrt {x}.

Now, \det \left(A^2 + A + x \mathbb{I}_{2 \times 2} \right) is the product of these eigenvalues, which is to say

    \[\det \left(A^2 + A + x\mathbb{I}_{2 \times 2} \right) = (i\sqrt{x})(-i\sqrt{x}) = x\]

as desired.

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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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