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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 eigenvectorsv_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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