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16

We have the matrix $$A= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$ Elininating the $a$s below the ...


13

Partition the matrix into blocks $\begin{bmatrix}A&B\\C&D\end{bmatrix}$, where $A = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, $B = \begin{bmatrix}0&0&a\\0&a&0\\a&0&a\end{bmatrix}$, $C = \begin{bmatrix}0&0&a\\0&a&0\\a&0&0\end{bmatrix}$, $D = ...


7

Hint: After rearranging rows, you get a matrix of the form $5 U-6I$ where $U$ is the all-ones matrix. Note that $U$ has rank $1$. Find the eigenvalues...


6

Hint: Let $$A:= \begin{bmatrix} 1 & 1 & 1 \\ 1 & \alpha & \beta \\ 1 & \alpha^2 & \beta^2 \end{bmatrix} $$ then calculate $AA^T$ to see what you get.


4

Note that $A$ (dimension $2\times 3$) has rank at most 2 and so cannot have full column rank. That is, there is $v$ ($3\times 1$) not identically $0$ such that $$ Av=0. $$ But then $(A'A)v=A'(Av)=0$, implying that $A'A$ cannot be invertible. Alternatively, note that $(A'A)v=0$ and $v\neq 0$ expose the fact that $A'A$ (a $3\times 3$ matrix) does not have full ...


3

Your error is in the step $$(4-\lambda)(\lambda^2 - 6\lambda - 91 + 75) = 0 \\ = (4-\lambda)(\lambda^2 - 6\lambda \color{red}{-21}) = 0$$ It should be $\color{blue}{-16}$ and then it factors well to give you the required answer


3

Let $A$ be a $m \times n$ matrix. The rank of $A^T A$ turns out to be the same as the rank of $A$, and $A^T A$ is a $n \times n$ matrix. So $A^T A$ is nonsingular if and only if the rank of $A$ is $n$, i.e. if $A$ has linearly independent columns.


3

Consider diagonal matrices with entries $\alpha$, $1,\ldots,1$ in the main diagonal. Then the only term in your sum is the one coming from the identity permutation.


3

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 ...


3

First of all, notice that exterior powers commute with base change (Eisenbud, Commutative Algebra with a View..., Proposition A2.2, p. 576), hence $$\Lambda^n_{F[x]} (F[x] \otimes_F V)=F[x] \otimes_F \Lambda^n_{F}V$$ You can easily check, that the following diagram (of $F$-modules) commutes ($m(\lambda)$ is the map $x \mapsto \lambda$ from the other ...


3

Partially yes, we can exhibit nine more or less "obvious" eigenvectors of eigenvalue $0$, namely $(k,0\ldots,0,-1,0,\ldots,0)^T$ where $-1$ is at the $k$th place, $2\le k\le 10$. So the kernel is indeed 9-dimensional. However, the eigenvalue $1$ should be wrong since $(1,2,\ldots,10)^T$ "clearly" is an eigenvector with eigenvalue $1^2+2^2+\ldots+10^2$.


2

The proof is invoking the uniqueness theorem for solutions to a Cauchy problem. We have the Cauchy problem $$y'(t) = Ay(t)$$ $$y(t_0) = 0$$ Now, we know there's only one solution to this. But clearly, the constant zero function is a solution, so therefore it's the only solution. Since your function $\sum c^* x^k$ is also a solution, by uniqueness, it ...


2

Fix a scalar $\lambda\in F$. There is an evaluation map $m_\lambda:F[x]\rightarrow F$ such that $x\mapsto \lambda$. It induces a commutative diagram $\require{AMScd}$ \begin{CD} \Lambda^n_{F[x]} M @>{\Lambda^n_{F[x]}(1\otimes T-x\otimes \id_V)}>> \Lambda^n_{F[x]} M\\ @V{m_\lambda}VV @V{m_\lambda}VV \\ \Lambda^n_F V @>{\Lambda^n_F (T-\lambda ...


2

Let us write $Te_i = \sum_j a^j_i e_j$. For $1 \leq i < j \leq n$, we have $$ (\Lambda^2(T))(e_i \wedge e_j) = Te_i \wedge Te_j = \left( \sum_{k_1} a_i^{k_1} e_{k_1} \right) \wedge \left( \sum_{k_2} a_j^{k_2} e_{k_2} \right) = (a_i^i a_j^j - a_i^j a_j^i) (e_i \wedge e_j) + \cdots $$ where the $\cdots$ don't involve $e_i \wedge e_j$ (as the coefficient ...


2

The determinant is : $$\begin{align}\begin{vmatrix} a&b&c\\\ b&c&a \\\ c&a&b \end{vmatrix}&=3abc-a^3-b^3-c^3 \\ &=-\frac 12(a+b+c)\left((a-b)^2+(b-c)^2+(c-a)^2\right) \\ &=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \end{align} $$ 1)This one implies $a=b=c\ne0$ so they are identical planes. 2)Here the determinant is non-zero so ...


2

Consider $\begin{pmatrix}\lambda& 1&0&0\\ 0&\lambda &1 &0\\ 0&0&\lambda& 1\\ 0&0&0&\lambda\end{pmatrix}$.The characteristic polynomial and the minimal polynomial are $(x-\lambda)^4$.


2

This is probably the shortest way:- $$ \begin{align} \begin{vmatrix} b^2c^2 & bc & b+c \\\ c^2a^2 & ca & c+a \\\ a^2b^2 & ab & a+b \end{vmatrix} &=a^3b^3c^3\begin{vmatrix}a^{-1} &1 & b^{-1}+c^{-1}\\\ b^{-1} &1&a^{-1}+c^{-1}\\\ c^{-1} &1& a^{-1}+b^{-1}\end{vmatrix}\\ &=(abc)^3\left(\frac 1a+\frac ...


2

In general $\text{rank}(AB) \le \max(\text{rank}(A), \text{rank}(B))$. In particular, if $A$ and $B$ are $m \times n$ and $n \times m$ with $n < m$, $AB$ must be singular. However, it's incorrect to say "the product of a matrix and its transpose, when they are not square, is singular": if $A$ is $m \times n$ with $n < m$ , $A A^T$ (which is $m ...


2

Hint see Vandermonde's determinant or write it as a multiplication of two determinants


1

I've found an easier way to accomplish what I want. Start with the integral symmetric matrix $M$. For $i=1, \ldots, n-1$ we can row-reduce each row using only scalar multiplication (of an integer) and adding a multiple of one row to the other. We row reduce each row until we have an upper triangular matrix. (That is, start by row-reducing to eliminate entry ...


1

This is a circulant matrix and as such has normalized eigenvectors $$v_j=\frac{1}{\sqrt{n}}(1,\omega_j,\omega_j^2,\ldots,\omega_j^{n-1})$$ where $$\omega_j=exp\left(\frac{2\pi i j}{n}\right)$$ The eigenvalues are $$\lambda_j=n+c\omega_j+c\omega_j^2+\cdots+c\omega_j^{n-1}$$ Taking the product of the eigen values gives the determinant. Now since the $\omega_j$ ...


1

Simply just use the formula for the determinant of a matrix you already know, ensuring you consider the arithmetic of complex numbers where necessary (for example, division by $a+ib$ occurring somewhere would require you to multiply through by conjugates etc.. The inverse would not exist is if the determinant of the matrix with complex entries is zero. If ...


1

Yes it is ; working in $\mathbb{R}$ or $\mathbb{C}$ does not change anything when dealing with determinant and inverses of matrices, though of course, the determinant of a complex matrice is a priori complex. As a reminder, I recall one of the definitions of the determinant (there are others, which are equivalent): ...


1

$F=\left| \begin{array}{cc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \\ \end{array} \right|$ $=\dfrac1{abc}\left| \begin{array}{cc} ab^2c^2 & abc & a(b+c) \\ c^2a^2b & bca & b(c+a) \\ a^2b^2c & abc & c(a+b) \\ \end{array} \right|$ $=\left| \begin{array}{cc} ab^2c^2 &1& a(b+c) ...


1

Your given matrix is : $\left| \begin{array}{cc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \\ \end{array} \right|$ Determinant of the given matrix is : $\implies b^2c^2[ca^2 + abc - abc + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$ $\implies b^2c^2[ca^2 + a^2b] - bc[a^3c^2 + ...


1

For $A\sim B$ you could calculate a new matrix $N_{ij} = \left\{\begin{array}{l}1 \text{ if } A_{ij}\ne B_{ij}\\0 \text{ if } A_{ij}= B_{ij}\end{array}\right.$, then define $A \sim B = \displaystyle\sum_{i=1}^n\sum_{j=1}^n N_{ij}2^{in+j}$. Then $\sim$ is basically a large binary number, each non-zero bit of which indicates a difference between $A$ and $B$ ...


1

$$\mathrm{tr}(\log (iA )) = \mathrm{tr}(i \pi/2 + \log A ) = i n \pi/2 + \mathrm{tr}(\log A )$$ When you multiply a matrix by a constant $c$, the determinant gets a factor of $c^n$. $$\log(\det ( i A )) = \log( i^n \det A ) = \log( i^n ) + \log( \det A ) = \log( i^n ) + \mathrm{tr}\log( A )$$ So in your case for the RHS to equal the LHS you want ...


1

hint Note that $a^2+b^2+c^2=ab+bc+ca$ is another way of saying $a=b=c$ (just multiply both sides by $2$ and complete the squares). So if (4) holds, then $a=b=c=0$, in which case the system is actually $0=0$, hence the solution space is all of $\mathbb{R}^3$. Try to proceed and see if you can complete now. Otherwise I can elaborate. Further elaboration: ...


1

The confusion probably stems from your retaining the original labelling of the entries of the matrix. For instance, in the formula for $\det [A]_1$, $a_{11}$ would refer to the first entry of $[A]_1$ which is actually called $a_{12}$. You probably conflated the two when you claimed the determinant did not change. Let $b_{ij}$ denote the entries of $[A]_1$ ...


1

$rank(M) = \min\{r|\exists V \in L(m,r), W \in L(n,r). M = V^TW \}$ where $L(m,n)$ is the space of $n\times m$ matrices is my rewriting of the (intended) meaning of your definition. You should verify that it is equivalent. I think this definition (particularly compared to your definition of rank as the dimension of the image) makes it much easier to answer ...



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