# Tag Info

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For any diagonalisable matrix $A$ whose distinct eigenvalues are $\lambda_1,\ldots,\lambda_k$ (meaning that this list contains each eigenvalue only once, regardless of the dimension of the corresponding eigenspace or of the multiplicity of each $\lambda_i$ as root of the characteristic polynomial), a polynomial $P$ annihilates $A$ is and only if each ...

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Such ansatz makes sure that your $A$ is still symmetric. If you just multiply by $M^{-1}$, you will not have a symmetric matrix on the right-hand side of the equation. Here's how the given ansatz works: \begin{align} M\frac{d^2}{dt}X & = KX \\ M^{1/2}\frac{d^2}{dt}\left(M^{1/2}X\right) & = KX \\ \frac{d^2}{dt}\left(M^{1/2}X\right) & = ...

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(An elementary proof without the use of the Cayley-Hamilton Theorem.) Clearly it suffices to show that the $i,j$-th entry of the $\text{RHS}$ $=C_{ji}$. We establish 2 simple facts. Fact 1: Let $\{k,m,i\}$ be a permutation of $\{1,2,3\}$. Then $$C_{ii}=a_{kk}a_{mm}-a_{km}a_{mk}$$ Proof: This follows from the definition of a cofactor and the formula ...

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You already answer the question. In summary $X'X$ is a positive-semidefinite matrix so eigenvalues are either $0$ or positive. When you say $X'X$ is positive-definite you are saying that eigenvalues are strictly positive so the matrix is full-rank.

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$$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right)$$

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Comment: The idea is that there are not enough "conditions" ("restrictions", "relationships") imposed to determine each variable uniquely. So you are free to assign values to unrestricted variables. For a more simple concrete example, suppose you are asked to find all solutions $(x,y)$ to the equation $x+ y=1$, where $x$ and $y$ are real numbers. You might ...

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$X\sim Y$ iff $XH=YH$ iff $XY^{-1}\in H$. If $X=\begin{pmatrix}p&q\\r&s\end{pmatrix},Y=\begin{pmatrix}u&v\\w&x\end{pmatrix}\in I$, then $X\sim Y$ iff $-pv+qu=0 \mod 4$. Let $H_1$ be the equivalence class of $X=I_2$: the condition on $Y$ is $v=0\mod 4$. Let $H_2$ be the equivalence class of $X=\begin{pmatrix}1&2\\0&1\end{pmatrix}$: ...

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$f(W)=constant-2tr(XH^TW^T)+tr(WHH^TW^T)$. The derivative is $Df_X:Z\rightarrow -2tr(XH^TZ^T)+tr(ZHH^TW^T)+tr(WHH^TZ^T)=$ $-2tr(XH^TZ^T)+2tr(ZHH^TW^T)=-2\langle XH^T,Z\rangle+2\langle WHH^T,Z\rangle$. Thus the gradient is $\nabla(f)(W)=-2XH^T+2WHH^T$.

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It's easy to see if you interpret the matrices in terms of linear maps. Let $u\colon K^n\to K^n$ the linear map associated to $A$, $v\colon K^n\to K^n$ the linear map associated to $B$. Then $u\circ v$ corresponds to $AB$. Now, it is always true that $\;\DeclareMathOperator\rk{rank}\rk(AB)\le\rk B$, so the hypothesis reallys means $\rk AB=\rk B$, in ...

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Hint: The easiest approach is to go to the definition of subset, i.e., show that every element of $P(B)$ is also an element of $P(AB)$. Now, if $x\in P(B)$, then $Bx=0$, so ...

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Since the rank of a product is less than or equal to the rank of each factor, the hypothesis $\operatorname{rank}(B)\le\operatorname{rank}(AB)$ implies $\operatorname{rank}(B)=\operatorname{rank}(AB)$. As a consequence, $$\dim P(B)=n-\operatorname{rank}(B)=n-\operatorname{rank}(AB)=\dim P(AB)$$ It is clear that $P(B)\subseteq P(AB)$ (if $x\in P(B)$, then ...

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If $A$ is nonsingular, then $AA^{-1} = I$, so $$1 = ||I|| = ||AA^{-1}|| \leqslant ||A||\cdot||A^{-1}||.$$ In general, then $1 \leqslant ||A||\cdot||A^{-1}|| \implies ||A||^{-1} \leqslant ||A^{-1}||$. Equality is thus not necessarily guaranteed for arbitrary nonsingular $A$; however, the inequality above implies that equality may occur. Consider an ...

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You can simply perform the product $XX^\top$ in the Boolean semiring (which amounts to use the rule $1+1 = 1$ instead of $1 + 1 = 2$). Then just change every $1$ by $0$ and every $0$ by $1$ in the resulting matrix.

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Let's go through an example of proving that one kind of elementary matrix does what it's supposed to. In particular, let $E$ be the elementary matrix that adds $\beta$ times the first row of an $n \times k$ matrix to the second row. We will know that $E$ does what it's supposed to do if, for an $n \times k$ matrix $A$, we have $$(EA)[i,j] = \begin{cases} ... 0 The key is to understand what matrix matrix multiplication really does. Consider the problem of computing the product C = AB, where all matrices are square of dimension n for the sake of simplicity. By definition we have $$c_{i,j} = \sum_{k=1}^n a_{ik}b_{k,j}$$ for each component c_{i,j}. If we zoom out and consider the i ... 1 The characteristic polynomial of a linear map is defined without any additional structure beyond that of a finite dimensional vector space while the notion of unitarity or normality requires the structure of an inner product so you should be suspicious if someone tells you that the characteristic polynomial tells you something about unitarity or normality. ... 3 Hint Consider the eigenvalues of A \in SO(3, \Bbb R). In particular, (1) their product is \det A = 1, (2) they all have modulus 1, and (3) any nonreal eigenvalues come in complex conjugate pairs. Now, if A \neq I, what can you say about the 1-eigenspace of A? 2 A direct argument may be obtained by arguing by induction. You should try the small cases n = 2, 3, 4 to see what this means in practice. Fix k. Say that a matrix B satisfies condition null-k if b_{ij} = 0 if i \ge j - k. (Clearly B = A satisfies null-0.) Suppose B satisfies null-k. Consider the product C = A B, and let c_{ij} be ... 2 Hint Recall that the adjugate matrix \DeclareMathOperator{adj}{adj} \DeclareMathOperator{tr}{tr} \adj A of an n \times n matrix A satisfies$$A \adj A = (\det A) I$$for all A. Now, if \lambda_1, \lambda_2, \lambda_3 are the eigenvalues of the 3 \times 3 matrix A, the characteristic polynomial of A is$$p_A(t) = (t - \lambda_1)(t - ...

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Recall that $A \cdot \text{adj}(A) = \det(A) I$. If you multiply the given equation by $A$, you get \begin{align} \det(A) I & = \frac 12\left[(\text{tr}(A))^2 - \text{tr}(A^2)\right]A - (\text{tr}(A))A^2 + A^3. \end{align} Now suppose $\lambda_1, \lambda_2, \lambda_3$ are eigenvalues of $A$. Then we can rewrite $\det(A)$, $\text{tr}(A)$ and ...

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This is obvious using Hamilton-Cayley's theorem: the only eigenvalue of such a matrix is $0$, since its characteristic polynomial is $\chi_A(x)=(-1)^nx^n$. By Hamilton-Cayley, $$\chi_A(A)=0=(-1)^nA^n.$$

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The rank is 2 (the first two columns span the image) and trace is zero. If there is one non-zero eigenvalue $a$, there is another non-zero eigenvalue $-a$. Hence the image would be spanned by the two eigenvectors of the non-zero eigenvalues and the same would automatically hold for all powers of $A$. But the first column is contained in the kernel and in the ...

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Since the characteristic polynomial is $\chi_A (x) = x^4$, this tells you that all eigenvalues are zero. A matrix satisfies its characteristic polynomial by the Cayley Hamilton theorem, hence $A^4 = 0$. If $v$ is a non zero eigenvector with eigenvalue $\lambda$ then $0 = A^4 v = \lambda^4 v$ and so $\lambda = 0$.

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As you may know, an interpretation could be as vectors and volumes in the dual space. But it is not very physically meaningful... I will give here a geometrical explanation in $\mathbb{R^3}$. Let $u,v,w \in \mathbb{R^3}$. Call $A$ the matrix with these columns $u,v,w$. Imagine the parallelotope $P$ generated by $u,v,w \in \mathbb{R}^3$. The volume of ...

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The fact builds upon the facts on eigenvalue and eigenvectors of symmetruc matrix. The one directly leads to the fact you asked is that: a symmetriic matrix A can decomposed as $$A = Q^TDQ$$ where Q is an orthogonal matrix and D is diagonal matrix and entries of diagonal of D is eigenvalues $\lambda_i$ of A: $d_{ii} = \lambda_i > 0$ Now $$x^TAx = ... 2 The straight-forward approach is Gram-Schmidt. Start with the standard (canonical) basis \{e_1,e_2\}. Our goal is to get an orthonormal basis \{v_1,v_2\}. First, we take$$ v_1 = \frac {1}{\sqrt{\langle e_1,e_1 \rangle}} e_1 = \frac 1 {\sqrt 2}e_1 $$Now, take$$ w_2 = e_2 - \langle e_2,v_1 \rangle v_1 = e_2 - \frac 1 2 \langle e_2,e_1 \rangle e_1 = ...

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(Assuming an algebraically closed field) the minimal polynomial is square-free iff $A$ is diagonalizable. This is most easily seen from the Jordan normal form, but can also be inferred directly by considering invariant subspaces per linear factor.

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User John Brevik gives as a hint: What is the characteristic polynomial of $$\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$$ How could you have come up with this hint yourself? Well, as you say, the characteristic polynomial only cares about eigenvalues, so you need to find two matrices which are not similar but which have the same eigenvalues. ...

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It is true in general that $$\operatorname{Tr}(AX)=\operatorname{Tr}(XA)\implies \operatorname{Tr}(x^T(By))=\operatorname{Tr}((By)x^T)$$

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We see this in quantum mechanics. We can expand linear operator in terms of bra and ket. $$A = \sum |x><x|A|y><y|$$ Then $\sum |x><x|=1$ is the identity matrix and $<x|A|y>$ are the matrix elements. Then again. Physics is know for suggestive notations which may not work 100% of the time.

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The quick answer to your question: note that the only diagonalizable matrix whose eigenvalues are all $0$ is the zero-matrix, and that a rank $1$ matrix can have at most one non-zero eigenvalue. Another approach: Note that any rank $1$ matrix can be written in the form $uv^T$ for column vectors $u,v$, and that a rank-$1$ matrix will be symmetric if and ...

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Let $\mathcal B=(e_1,\ldots,e_n)$ be a basis of $V$ and $\mathcal B^*=(e_1^*,\ldots, e_n^*)$ the associated dual basis of $V^*$. The map $$V\otimes V^* \rightarrow End(V)$$ sends $e_i\otimes e_j^*$ to the linear operator : $$v=(v_1,\ldots,v_n)\mapsto e_j^*(v)e_i=\sum_{k=1}^nv_je_i.$$ Hence, the matrix representation of $e_i\otimes e_j^*$ in the basis ...

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The minimal polynomial of $A$ divides $x^4 - 4x^2 = x^2(x-2)(x+2)$ and the characteristic polynomial and the minimal polynomial have the same irreducible factors (possibly with different multiplicities), so a priori, you can only tell that the possible eigenvalues are $0,\pm 2$ and not necessarily all must occur (for example, the matrix $A = cI$ where c \in ... 4 Caley-Hamilton gives you $$A^4 - 2A^2 + 1 = 0 \quad \Leftrightarrow \quad A^4 = 2A^2 - I \quad \Rightarrow \quad A^8 = (2A^2 - I)^2$$ EDIT: As Michael suggested, this can be further simplified such that $$A^8 = 4A^4 - 4A^2+ I = 4(2A^2 - I) - 4A^2 + I = 4A^2 -3I$$ 1 In your case you can write \begin{align*} M&=(2e_1+5e_2)\otimes e_1+(3e_1+7e_2)\otimes e_2\\ &=2e_1\otimes e_1+5e_2\otimes e_1+3e_1\otimes e_2+7e_2\otimes e_2. \end{align*} The idea is the following: choose a basis\{x_1,\ldots,x_k\}$of the range of the operator$T.$Then write $$Tx=\lambda_1(x) x_1+\cdots+\lambda_k(x) x_k$$ for some scalars ... 0 Use the minimal polynomial, which is$\;(x+4)(x-5)\;$: $$0=(A+4I)(A-5I)=A^2-A-20I\implies 20I=A^2-A\implies A^{-1}=\frac1{20}\left(A-I\right)$$ 2 Hint: the minimal polynomial is as you said$x^2-x+20$. Then, $$A^2-A-20I = 0$$ $$A(A-I) = 20I$$ $$\dots$$ 1$n=2$,$\newcommand{\tr}{\operatorname{tr}}$$$\det(A+B) = \det(A) + \det(B) + \tr(A) \tr(B) - \tr(AB).$$$n=3$, letting$c(X) = (\tr(X)^2 – \tr(X^2)) / 2, \begin{align*} \det(A + B) ={}& \det(A) + \det(B) – \tr(AB)\tr(A) – \tr(AB)\tr(B) +{} \\ &{}+ c(A)\tr(B) + \tr(A)c(B) + \tr(AAB) + \tr(ABB) \end{align*}n>3$, a formula with$2^n$... 0 The matrix is diagonalizable for the spectral's theorem. Indeed the dimension of the eigenspace for the eigenvalue of$0$is$n-1$. Note that the eigenspace for the eigenvalue of$0$is the$\ker$of function. 0 Consider two matrix: $$A=\begin{bmatrix} 1 & 0 & -19 \\\\ 0 & 1 & -7 \\\\ 0 & 0 & k-26 \end{bmatrix}$$ And $$B=\begin{bmatrix} 1 & 0 & -19 & 2\\\\ 0 & 1 & -7 & -1\\\\ 0 & 0 & k-26 & h+1 \end{bmatrix}$$. The linear system has solutions when$rk A=rk B$(for the Rouché-Capelli's theorem) ... 1 Indeed, the only requirement to have the inequality that you wrote for the Frobenius norm and for arbitrary matrices is that the product$AB$is defined. If you are looking for a reference see for example the book Numerical Linear Algebra, by Trefethen and Bau, page 23. 1 Normal means precisely that$AA^\ast =A^\ast A$. Then $$\langle Av,Av\rangle=\langle v,A^\ast Av\rangle=\langle v,AA^\ast v\rangle=\overline{\langle AA^\ast v,v\rangle}=\overline{\langle A^\ast v,A^\ast v\rangle}$$ This gives the first result. Can you deduce the second from the first? 3 Since$N$is nilpotent, there is a$r\in \Bbb{N}$that$N^r=0. By the fact that $$(I-N)(I+N+\cdots +N^{r-1})=I-N^r=I$$ We conclude $$(I-N)^{-1}=I+N+\cdots +N^{r-1}\tag1$$ In your example $$D=\pmatrix{2 \\ &2 \\ && 3}\quad\text{and }\quad D^{-1}=\pmatrix{1/2 \\ & 1/2 \\ && 1/3}$$ And N=\pmatrix{0 & -1 & -2 \\0 & ... 2 Here's an attempt to make the 2\times 2 case seem just slightly horrible, while practicing to master the align* environment. Let A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} and B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}. Then the left-hand side expands to eight terms: ... 0 Suppose A is nonzero, otherwise the statement is trivial. As A is 3\times3 and skew-symmetric, it is a cross-product matrix [ru]_\times for some r\in\mathbb R and unit vector u. Extend u to an ordered orthonormal basis \{u,v,w\} so that u\times v=w. So, (AB+BA)x=0 for x\in\{u,v,w\} if and only if \begin{align} u\times Bu &= ... 0 Every one of those matrices is just the original with its rows permuted. As I mentioned in my comment, left-multiplying a matrix by the ith row of the identity picks out its ith row. For example, \left[\begin{matrix}0&1&0\end{matrix}\right] \left[\begin{matrix}a&b&c\\d&e&f\\g&h&i\end{matrix}\right] = ... 0 Letv=\begin{bmatrix}1\\1\\0\end{bmatrix}$Let$\Delta = \begin{bmatrix}1&1&1\\1&1&1\\0&0&0\end{bmatrix}A\cdot \Delta = A\cdot \begin{bmatrix}1&1&1\\1&1&1\\0&0&0\end{bmatrix} = \begin{bmatrix}Av&Av&Av\end{bmatrix}$In particular, all columns of$A\cdot \Delta$will be identical for any$A$. As ... 0 Left multiply the original matrix by the following rearranged identity matrix, and see what happens. You should be able to get an idea of the purpose of this exercise. \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{matrix} 0 Your assumption is correct! Here "null" means a row vector of length$n$with each component equal to zero. Personally, I would have written the matrix as $$A = \begin{pmatrix} \alpha & 0^T \\ v & B \end{pmatrix},$$ but that is because I prefer to deal with scalars, and column vectors rather than row vectors. By expanding ... 7 In the cases of$2 \times 2$matrices it always holds. This appears to be a horrid calculation so by introducing a variable$\lambda$$$\text{det}(\lambda A + B)-\lambda^2 \text{det}(A) - \text{det}(B) - \lambda \text{Tr}(A)\text{Tr}(B) + \lambda \text{Tr}(AB)$$ is a polynomial. To get the quadratic part of$\text{det}(\lambda A + B)\$, write ...

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