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0

Let the two vectors in $B$ be $\vec v_1$ and $\vec v_2$. Then $\underset{B \leftarrow A}{P}$ (the change of basis matrix from $A$ to $B$) can be found by forming the augmented matrix $[B \mid A]$ and row-reducing to $[I \mid \underset{B \leftarrow A}{P}]$. Alternatively, we can notice by inspection that: \begin{align*} \vec v_1 + \vec v_2 = -2\vec e_1 \\ ...


0

Let me write $F_5$ for the field of fivve elements. Now, suppose $x_1,x_2,\dots,x_r$ are all solutions of $Ax=0$, and suppose $c_1,c_2,\dots,c_r$ are all elements of $F_5$. Then $c_1x_1+c_2x_2+\cdots+c_rx_r$ is also a solution, because $$A(c_1x_1+c_2x_2+\cdots+c_rx_r)=c_1Ax_1+c_2Ax_2+\cdots+c_rAx_r=c_10+c_20+\cdots+c_r0=0$$ So, if the equation has one ...


0

Since $P$ is supposed to have a right inverse, it must necessarily have rank $m$. For a nontrivial answer, assume $Px\neq 0$. A particular example of $\bar P$ can easily be constructed as follows. Consider the Moore-Penrose pseudo-inverse $P^+$ of $P$ so that $PP^+=I$ and note that the range of $P^+$ is orthogonal to the nullspace of $P$. Consequently, we ...


5

Our linear map is $$ \begin{array}{ccccc} T & : & \mathcal M_{2\times 2}(F) & \to & F \\ & & A & \mapsto & \DeclareMathOperator{tr}{tr}\tr(A) \end{array} $$ The basis $\alpha$ for $\mathcal M_{2\times 2}(F)$ is $$ \alpha = \left\{ \begin{bmatrix}1 & 0\\ 0 & 0 \end{bmatrix}, \begin{bmatrix}0 ...


0

You are going about it in the right way, but one thing you should do is to always keep denominators real. Thus for example $-\frac{1}{i} = i$ and $\frac{1}{4+i} = \frac{4-i}{(4-i)(4+i)} = \frac{4-i}{17}$. So your next steps would be \begin{align*} \sim \left(\begin{array}{ccc|c} 1 & -2 & i & 0 \\ 0 & 1 & \frac{-1+3i}{i+4} & ...


0

I believe what you are looking for is the Moore-Penrose inverse. If your matrix has full rank, then $PP^T$ is an invertible $m \times m$ matrix, and the Moore-Penrose inverse is $\bar P = P^T(PP^T)^{-1}$. The matrix $\bar P$ is an $n \times m$ matrix such that $$\bar P|_{\operatorname{range}(P)}: \operatorname{range}(P) \to \ker(P)^\perp$$ is an isomorphism. ...


1

All that I can interprete into your $E$ is that it ought to be a linear map $\mathfrak{sl}(2,\mathbb K)\to\mathfrak{sl}(2,\mathbb K)$. It looks like the Lie bracket with $h$ perhaps: $x\mapsto[h,x]$.


2

There is no need to bring any outside information about Pell's equation. Just take some non-trivial unit, say $x = \begin{pmatrix} 3 & 4 \\ 2 & 3 \\ \end{pmatrix}$, and observe that elements $1,x,x^2,x^3,\ldots$ are distinct. In fact, the $x$ generates the group of units, but that is not necessary here.


2

If $a=3$ and $b=2$, then $a^2-2b^2=1$. In particular $(3+2\sqrt{2})(3-2\sqrt{2})=1$. Consider $$ x_n=(3+2\sqrt{2})^n=a_n+b_n\sqrt{2},\quad y_n=(3-2\sqrt{2})^n. $$ What can you say about $a_n$, $b_n$ and $a_n^2-2b_n^2$? (Use that $x_ny_n=1$). You can note that the map $$ \begin{bmatrix}a & 2b \\ b & a\end{bmatrix} \mapsto a+b\sqrt{2} $$ is a ring ...


1

The formula $\dim(\operatorname{Ker} T)+\dim(\operatorname{Im} T)=\dim V=n$ from the "rank + nullity theorem" should be enough for this. If $T$ is injective, then $\dim( \operatorname{Ker} T) =0$ so $\dim(\operatorname{Im} T) =n$, therefore since $\operatorname{Im} T$ is a subspace of $V$ you get $\operatorname{Im} T=V$, because the only subspace of $V$ ...


0

I think it is essentially correct, but I would write things a bit differently. You are right on using rank-nullity. But then I would say that the only vector $z$ with $T(z)=0$ is the zero vector. Then the kernel is 0-dimensional, i.e., nullity is $0$, so rank is $n$, and then $T(V)$ is an n-dimensional subspace of $V$, and so it must be the whole of $V$ ...


1

The first part is okay, but the second one (finding $T(1)$) isn't. Note that linear transformations give you $$T(1) = -T(-2x^2 + (2x^2 - 1)) = -T(-2x^2) - T(2x^2 - 1) \\ = -(3x^2 - 3x) - (-3x +2) = -3x^2 - 2$$ With your approach, you'd have to find that $$1 = -1\cdot (-2x^2) + 0\cdot (\frac12 x + 4) - 1\cdot(2x^2 - 1) = \pmatrix{-1\\0\\-1}_{\mathcal B}$$ So ...


1

The kernel of the matrix is a vector subspace of $(\mathbb{Z}/5)^9$. So, it must be isomorphic to $(\mathbb{Z}/5)^k$ for some $0\leq k\leq 9$. The rank of the matrix is the dimension of the range minus the dimension of the kernel.


0

$T(1,1)$ and $T(1,2)$ should be expressed in basis $\mathcal B'$, not $\mathcal B$. Let $\{e_1, e_2\}$ be the canonical basis. Set $u_1=\begin{bmatrix}1\\1 \end{bmatrix}=e_1+e_2$, $\,u_2=\begin{bmatrix} 1\\2\end{bmatrix}=e_1+2e_2$. From these, you deduce: $$e_1=2u_1-u_2,\quad e_2=u_2-u_1.$$ You've proved $\,T(u_1)= 3e_1-2e_2$, $\,T(u_2)=4e_1-5e_2$, ...


2

The way you have things defined, it looks like you can read off the column vectors you seek from that matrix with respect to the linear transformation. The matrix of $L$ with respect to $\mathcal{S}$ and $\mathcal{T}$ is one that is precisely $$\begin{bmatrix}[L\mathbf{v_1}]_{\mathcal{T}} & [L\mathbf{v_2}]_{\mathcal{T}} & ...


0

A parallelogram with side lengths $a$ and $b$, with angle $\theta$ between them always has area $$ab\sin\theta,$$ regardless of the dimension in which the parallelogram lives. Here, $a = \lVert {\bf a} \rVert$, $b$ is defined analogously, and you can find $\theta$ using the standard technique involving the dot product. As you're seeing, there are many ...


0

Form a matrix $A$ whose columns are the given vectors. Then $\det (A^TA)$ is the square of the area of the parallelogram. This generalizes easily to $k$ vectors in ${\Bbb R}^n$.


0

Take the cross product. Or, more generally for vectors in $\Bbb R^n$, take the wedge product. http://en.wikipedia.org/wiki/Exterior_algebra


1

Depending on how you define the area (and or the region), the proof techniques might differ, but in the end the idea is always the same : the determinant of the matrix is the multiplication factor for the area. In other words, if $K$ has area $\lambda$, then $BK$ has area $|\det B| \cdot \lambda$. I assume you can compute the determinant. As for the proof, ...


2

$$\begin{cases} 0x+0y=0 \\ 0x+\frac 32y=0\end{cases} \iff \begin{cases} 0=0 \\ \frac 32y=0\end{cases}$$ That first equation doesn't give us any information. It's simply a tautology (something of the form $a=a$). Solving the second equation however, we see that $y=0$. But you can see that there aren't any conditions on the value of $x$. So the solutions ...


1

are you trying to find the eigenvalues and eigenvectors of the diagonal matrix $$A = \pmatrix{1/2&0\\0&2}? $$ we have $$Ae_1 = \frac 1 2 e_1, Ae_2 = 2e_2 \text{ where } e_1 \ \pmatrix{1\\0}, e_2=\pmatrix{0\\1}.\tag 2$$ the equation already tells you that the eigenvalues of $A$ are $1/2, 2.$


1

The condition $$ \frac{3}{2}y=0 $$ gives $y=0$. However, there are no more conditions and so $x$ may take any real value.


0

If the matrix (in general, let's say $n \times n$) is $A$ and the initial population vector ${\bf v}$, after $k$ generations the population vector is $A^k {\bf v}$. If there are $n$ linearly independent eigenvectors ${\bf u}_j$ and $\lambda_j$, we have ${\bf v} = \sum_j c_j {\bf u}_j$, and $A^k {\bf v} = \sum_j c_j \lambda^k {\bf u}_j$. If all ...


1

For the eigenvalue $\lambda=0$, that means that for the corresponding eigenvector, $x$, you have $Ax = \lambda x$ which in this case corresponds to $Ax=0$. To solve for $x$, you can set up and row reduce the augmented matrix: $\left[\begin{array}{ccc|c} 0 & -1 & -1 & 0\\ 0 & -1 & -3 & 0\\ 0 & 0 & -2 & ...


1

We have $$\begin{bmatrix} 0 & -1 & -1 \\ 0 & -1 & -3 \\ 0 & 0 & -2 \end{bmatrix} \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = 0\cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$ So the column vector is indeed an eigenvector of the matrix, with eigenvalue zero. I'm not sure what you ...


1

Your matrix is the sum between an identity matrix and a circulant matrix, so the characteristic polynomial is given by: $$ p(\lambda)=(1-\lambda)^n-(-1)^n \tag{1}$$ and the determinant is given by $(-1)^n p(0)$, so it is $2$ if $n$ is odd and zero otherwise.


0

In $\mathbb{Z}_2$ the system matrix has determinant $0$, which means that the solution cannot be unique. Therefore, let us look at only the first $n-1$ equations, with $x_n$ considered not as an unknown, but as a parameter:then you solve the system backwards and get the solution $x_1 = p_n + x_n$ (remeber that in $\mathbb{Z}_2$ $+$ and $-$ coincide), $x_2 = ...


1

The general matrix is given by the sum between the identity matrix and a circulant matrix, hence its characteristic polynomial over $\mathbb{Q}$ is given by: $$ p(\lambda)=(\lambda-1)^n-1.$$ Over $\mathbb{F}_2$ such a matrix cannot be invertible since the sum of the elements in every row/column is zero, hence $(1,1,\ldots,1)$ is an eigenvector associated ...


1

Of course, your conjecture is false. For instance, take $U=Rot(\theta),z=[\cos(\alpha),\sin(\alpha)]^T,y=[\cos(\alpha),-\sin(\alpha)]^T$. Then $|z|^2=|y|^2$ and $|Uz|^2=[\cos^2(\theta+\alpha),\sin^2(\theta+\alpha)^T$ and $|Uy|^2=[\cos^2(\theta-\alpha),\sin^2(\theta-\alpha)]^T$. Keep your bounty for further opportunity.


0

All you have to do is to find the $A^{\infty} = \lim\limits_{n \rightarrow +\infty} A^n$ and see what happens when you apply it to arbitrary "distribution" of pie and cake lovers. However, check your eigenvalues: this matrix clearly has one eigenvalue equal to 1 (I vaguely recall, but I'm not sure that all matrices which columns sum to 1 have this ...


0

If $\lambda=1$ is an eigenvalue, then there is a vector $r$ such that $Ar=r$ Let $r=xi+yj+zk$ $Ar=r$ would mean that $3x-2y+3z=x$ or $2x-2y+3z=0$ $-y+3z=y$ or $3z=2y$ $-x+2y-2z=z$ or $-x+2y-3z=0$ Are these equations consistent?


2

$$A-I=\begin{bmatrix} 3 & -2 & 3\\ 0 & -2 & 3\\ -1 & 2 & -3 \end{bmatrix}$$ And you can see the two last columns are proportionate so the matrix is not invertible and $\lambda=1$ is an eigenvalue My computation of $\det(A-\lambda I)=-\lambda^3+\lambda^2+13\lambda-13$


2

no matter if $\epsilon \neq 0$ is , there is no real matrix $M.$ for $\epsilon = 0, M = \pmatrix{0&b\\-\frac 1b&0}, b\neq 0$ pick a matrix $$M = \pmatrix{a&b\\c&d}, M^2 =\pmatrix{a^2 + bc&(a+d)b\\(a+d)c&bc+d^2}=\pmatrix{-1&0\\0&-1-\epsilon} $$ we have the constraints $$a^2 + bc = -1, bc+d^2 = -1-\epsilon \to a^2 - d^2 = ...


1

Assume $v$ is an eigenvector of $M$ with eigenvalue $\lambda$. Then $v$ is eigenvector of $M^2$ wih eigenvalue $\lambda^2\ge0$. Since $M^2$ has only negative eigenvalues $-1$ and $-1-\epsilon$, $M$ has no eigenvectors. Thus the matrix $A$ that maps $e_1\mapsto e_1$ and $e_2\mapsto Me_1$ is invertible and since $M$ maps $Me_1$ to $-e_1$ we find that ...


2

Let $M=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$. Then we want $M^2=\left(\begin{matrix}a^2+bc&b(a+d)\\c(a+d)&d^2+bc\end{matrix}\right)=\left(\begin{matrix}-1&0\\0&-1-\epsilon\end{matrix}\right)$. Now we can't have $b=0$ or $c=0$ (why?). Therefore $a=-d$ and ...


3

(I assume $M$ is supposed to be real; if it is allowed to be complex, then the exercise is trivial.) Hint What are the possible eigenvalues of $M$? What can one say about the eigenvalues of real $2 \times 2$ matrices...?


1

$$\begin{pmatrix} 1 & -2 & 3 & 2\\ 3 & 2 & 1 & -4\\ -1 & 6 & -7 & -7\\ -2 & 0 & -2 & 1 \end{pmatrix}$$ is a zero divisor since : column$(3)=$column$(1)-$column$(2)$. So the matrix: $$B=\begin{pmatrix} 0 & 0 & -a & 0\\ 0 & 0 & a & 0\\ 0 & 0 & a & 0\\ 0 & 0 & 0 ...


3

I assume you are working with $n \times n$ matrices over the complex numbers. Let $A^\ast$ denote the Hermitian adjoint of $A$, i.e. the complex conjugate of the transpose. Then, by definition, $A$ is Hermitian if and only if $A = A^\ast$. Now suppose that $A$ is an arbitrary complex $n \times n$ matrix. Set $B = \frac{1}{2}(A + A^\ast)$ and $C = ...


3

by row reducing $A-3I,$ using my ti-83, i get $$\pmatrix{1&-2&3&2\\3&2&1&-4\\-1&6&-7&-7\\-2&0&-2&1}\to\pmatrix{1&0&1&-0.5\\0&1&-1&-1.25\\0&0&0&0\\0&0&0&0} $$ we can see that two linearly independent vectors $a, b$ such that $(A-3I)a = 0, (A-3I)b = 0$ where ...


0

Let $A=\left(a_{ij}\right)$ a $n$ by $n$ real matrix. By definition, $$\det\left(A\right) :=\sum_{\sigma\in\mathfrak{S}_n}\varepsilon\left(\sigma\right)\prod_{j=1}^{n}a_{j\sigma\left(j\right)}$$ where $\mathfrak{S}_n$ is the group of permutations of $\{1,\ldots,n\}$ and $\varepsilon$ is the signature. The multilinearity of the determinant has to be shown on ...


3

One way to prove this is through eigenvalues. Since $A$ is positive definite, it is a symmetric matrix. $$A \text{ is positive definite } \iff \text{ all eigenvalues are positive. }$$ It is known that for any matrix $M$: $$\lambda \text{ eigenvalue of matrix } M \implies \lambda^k \text{ eigenvalue of matrix } M^k,\quad k=1,2,\ldots$$ Also, you can use ...


0

This is one of those many occasions where a more abstract approach will facilitate understanding. What does is mean that a linear operator $\phi$ has a diagonal matrix with respect to a given basis$\def\B{\mathcal B}~\B$? Answer: that every vector of$~\B$ is an eigenvector of$~\phi$. So the diagonal matrix $D_1$ represents a linear operator on$~K^n$ (with ...


0

If $A$ is the permutation matrix corresponding to the permutation $\pi$, and $B=(b_{ij})$, then $AB=(b_{\pi(i)j})$ and $BA^T=(b_{i\pi(j)})$, thus $ABA^T=(b_{\pi(i)\pi(j)})$. A diagonal matrix $D$ is defined by $i\ne j\implies d_{ij}=0$. Assume $i\ne j$. We have for the elements of $D'$: $d'_{ij} = d_{\pi^{-1}(i)\pi^{-1}(j)}$ Since a permutation is a ...


1

(1) Consider the map $$ \rho : S_n \rightarrow M_n(\mathbb{R}^n) $$ where $S_n$ is symmetric group. So if $A$ is a permutation, then it is an image of $\rho$. Let $\rho(t)=A$. And note that if $t=t_1\cdots t_m$ where $t_i$ is a single permutation, then $$ A_i:=\rho(t_i),\ A=A_1\cdots A_m$$ To show that $ADA$ where $D$ is diagonal, we suffice to show that ...


1

Try using the definition of semi-positive ($v^*Av\geq0$), the fact that all $x_i\in[0,1]$ (you need both bounds!) and Young's inequality ($2ab<a^2+b^2$) for $n$ terms. A 0-eigenvector (which is a slick way to say solution to the homogeneous system) is perhaps written in front of you?


0

One way is to first expand the right summand in the first equation - i.e. perform the vector-matrix multiplications. Then modify the result using for example the identity given here: Inverse of a sum of symmetric matrices. Further modify using the matrix inversion theorem (here a special case: $U=V=I$) - it's actually shown on the next page of one of the ...


0

in layman terms , the explanation is as follows. Lets say you have a matrix A and you have to reduce it To U(reduced echelon form) by using the operations mentioned in the pdf. Now lets say ,during this process ,you have a row where the elements have a common factor k1 .So as per the property you take the k1 constant outside the matrix i.e k1 [matrix]. Now ...


0

Just to answer my own question with a slightly alternative interpretation of the result as given by Alamos, in case it might be of some use to others in future: Let $P$ be a permutation matrix corresponding to elementary column operations that will move all zero columns in $N$ to the left. Then $NP=[0,X_1]$ with $X_1$ full column rank. Now $P^{-1}$ will also ...


0

The determinant is a multilinear function of the rows. Now $kR_i\to R_i$ amounts to multiply the $i$-th row, hence the determinant, by $\,\dfrac1k$.


2

(Cut-and-pasted from my comment above, per request of OP) "Matrices" with zero rows/columns can be thought of as linear transformations to/from zero-dimensional vector spaces. A $0\times n$ matrix can also be thought of as an empty array of numbers which maps everything to the zero vector. They have their usefulness in specifying initial/terminal objects in ...



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