# Tag Info

3

If $T(A)=\lambda A$, then $T^2(A)=\lambda^2 A$. But $T^2=I$ which only has an eigenvalue of 1. So $\lambda^2=1$ or $\lambda=-1,1$.

3

It means that $$\forall j \quad |\lambda_j(A)|<1$$ where $\lambda_j(A)$ are eigenvalues of $A$. There're several different ways of showing that $$\lim_{k\to\infty}A^k=0.$$ The first method is based on the Jordan normal form of the matrix (this form helps to find another expression for $A^k$). Essentially we show that any matrix is similar to an ...

3

$x$ is an eigenvector of $A$ if $x\neq 0$ and there exists $\lambda$ such that $Ax=\lambda x$. $\lambda$ is then called an eigenvalue of $A$ associated to the eigenvector $x$. Note in particular that every vector $x\neq 0$ in the kernel of $A$ is an eigenvector of $A$ associated to the eigenvalue $0$. More generally $\ker(A-\lambda I)$ is the space of ...

3

$$\det(I+ x A) = x^n \det (A + \frac 1 x I) = x^n\left(\lambda_1 + \frac 1 x\right)\left(\lambda_2 + \frac 1 x\right)\cdots \left(\lambda_n + \frac 1 x\right)$$

2

Eigen vectors corresponding to distinct eigenvalues are necessarily independent.Also $x_2$ can't be an eigenvector since it is $0$. When you solve, you find the eigen space $E_2$ has dimension $1$, with basis $x_1$. You complete it in a basis of the generalised eigenspace, which happens to be $\ker(A-2I)^2$. For instance, you may solve for ...

2

All of the vectors you give belong to $\lambda = -1$. This set is not linearly independent; the rank of $$\begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & -1 \\ 1 & -1 & 0 \end{pmatrix}$$ is 2, which means each two of them are sufficient to span the 2-dimensional eigenspace for this eigenvalue. The eigenvector for $\lambda = 2$ is $$... 2 What you want is |cY|=|c|\cdot |Y| \le 1; knowing that |Y|\le 1 and that the boundary can be reached, it is necessary and sufficient to have |c|\le 1. So your second answer, [-1,1], is correct. 2 It suffices to prove that$$2xx^T-2 \textrm{diag}(xx^T)+\mathbb{I}_n\geq 0$$or equivalently that$$\left[\matrix{1 & 2x_1x_2 & \cdots & 2x_1x_n\\ 2x_1x_2 & 1 & \cdots & 2x_2x_n\\ \vdots & \vdots & \ddots & \vdots\\ 2x_1x_n & 2 x_2x_n & \cdots & 1}\right]\geq 0$$Let now e_i the i-th column of ... 2 In Step 2 to Step 3, t-5 is factored out of the first row and the third row. The multiplies the determinant by the same factor, each time the factoring is done. In Step 4 to Step 5, expansion by minors is done on the first column. This is easy since there is a 1 in only one position and 0's in the other positions in that column. What's more, adding ... 2 \det(L-\lambda I)=-\lambda\begin{vmatrix}-\lambda & 0\\\frac{1}{3} &-\lambda\end{vmatrix}-\frac{1}{2}\begin{vmatrix}\frac{3}{2}a^2 & \frac{3}{2}a^3\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{1}{2}(\frac{3}{2}a^2)\begin{vmatrix}1&a\\\frac{1}{3}&-\lambda\end{vmatrix}=-\lambda^3-\frac{3}{4}a^2(-\lambda-\frac{1}{3}a) ... 2 T is a linear operator. As I understood this: find \lambda that:$$T(A) - \lambda A = 0$$If A is symmetric A = A^t then \lambda = 1, if anti-symmetric A = -A^t then \lambda = -1. In general case, there is no \lambda that satisfy a_{ij} = \lambda a_{ji} for all a_{ij} 1 Let's use the "brute force" approach here. We have the matrix equation Av=\lambda v. Thus, the eigenvalue equation becomes$$(A-\lambda I)v=0$$which implies that the determinant of A-\lambda I is zero. Thus, we have$$(\lambda -a)(\lambda -d)-bc=0$$which implies that$$\begin{align} \lambda &=\frac{(d+a)\pm \sqrt{(d+a)^2-4(ad-bc)}}{2}\\ ...

1

In chepukha's statement there is a small mistake. The radius must be $r_i=\sqrt{\frac{k}{\lambda_i}}$ (in his statement there is a square root missing, or, alternatively, the ellipsoid should be defined as $\sum_i\sum_j (x_i-u_i)(x_j-u_j)c_{ij}\leq k^2$). For example, consider the scalar case (where the proof is straightforward): equation $x^2 c \leq k$ has ...

1

Your work is correct . You are calculating a determinant and in the step $2 \rightarrow 3$ is used the fact that a determinant is a multilinear function of the rows ( and of the columns) of the matrix. The same is obtained using the fact that the determinant of a product is the product of the determinants. Explicitly: $... 1 Let$A \in \mathbb{R^{m \times n}}$and$\lambda$be non-zero eigen value of$AA^TAA^T \in \mathbb{R^{m \times m}}$and$A^TA \in \mathbb{R^{n \times n}}$$$AA^T\textbf{x} = \lambda \textbf{x}$$$\textbf{x} \in \mathbb{R^{m}}$Multiply(pre) both sides by$A_{n \times m}^T$$A^T(AA^T)\textbf{x} = \lambda A^T\textbf{x}$$ Rearranging a bit: ... 1 If you have eigenvector and eigenvalue pairs \begin{align*} v_1 &= (2, 1, 0)^T, &\lambda_1 = 1 \\ v_2 &= (-1, 0, 1)^T, &\lambda_2 = 1 \\ v_3 &= (1, -2, 1)^T, &\lambda_3 = -1 \end{align*} I would expect this to be a reflection, rather than a rotation. There are two reasons this would be so. First, the determinant is-1 = 1^2 ...

1

Compute $$\begin{pmatrix}-2 \\1 \\ 1 \end{pmatrix} - \begin{pmatrix}1 \\-2 \\ 1 \end{pmatrix} + 3\begin{pmatrix}1 \\-1 \\ 0 \end{pmatrix}$$

1

Translating result appeared on the wiki page of Characteristic polynomial, $$\det(I_n + x A ) = \sum_{k=0}^n x^k \text{tr}(\wedge^k A)$$ where $\text{tr}(\wedge^k A)$ is the trace of the $k^{th}$ exterior power of $A$ which can be evaluated explicitly as the determinant of the $k \times k$ matrix, \text{tr}(\wedge^k A) = \frac{1}{k!} ...

1

Hints: 1) The matrix's characteristic polynomial is $\;x^4\;$ (i.e., the matrix is nilpotent) 2) A matrix is diagonalizable iff its minimal polynomial is the product of different linear factors 3) Thus, a nilpotent matrix is diagonalizable iff ...

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