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5

Pick $F$ to be some field, and $X$ to be some uncountable set. Consider now $V$ to be the vector space whose elements are functions $f\colon X\to F$ such that all but finitely many $x\in X$ satisfy $f(x)=0$. The addition is pointwise addition, and scalar multiplication is pointwise multiplication. One can easily check that this is a vector space. Now it is ...


0

Any alternating $(r,s)$ tensor has a corresponding map that goes $\Lambda^r V \to \Lambda^s V$. Suppose $R \in \Lambda^r V$ and $\Sigma \in \Lambda^s V^*$. Then define $\underline T:\Lambda^r V \to \Lambda^s V$ such that $$T(R, \Sigma) = \Sigma[ \underline T(R)]$$ The uniqueness of $\underline T$ can be proved by taking a "gradient" with respect to the ...


1

In addition to the inequality mfl gave, there is a rather intuitive way of answering this question. The trace is the sum of all eigenvalues (with multiplicities): $\text{tr}(A) = \sum_i \lambda_i$. The 2-norm of a symmetric matrix is the maximum eigenvalue: : $||A||_2 = \max_i |\lambda_i|$. If you have negative and positive eigenvalues, the trace may ...


3

This is a parallelogram with edges $y = \alpha x - \frac{1}{N}, y = \alpha x + \frac{1}{N}, x = -N - \frac{1}{2}, x = N + \frac{1}{2}$, hence is convex with area $\frac{2}{N}((N + \frac{1}{2}) - (-N - \frac{1}{2})) = 4 + \frac{2}{N}$.


0

Here is a trivial operator: Ker F is a subspace so it will have a basis $u_{1},...,u_{k}$. Then extend it to basis $u_{1},...,u_{k},w_{1},...,w_{n-k}$ for V (dim(V)=n). Also, let f be a basis element of W. Then define: $F(v):=\left\{\begin{matrix}0 & if v\in Ker F\\ f & v\notin Ker F \end{matrix}\right.$ F in terms of T. $F(v):=0$ if $v\in$ ...


1

If the matrix $A$ is real symmetric it is diagonalizable. If $v$ is a unit eigenvector and $\lambda$ the corresponding eigenvalue, then $$|\lambda|=|\lambda v|_2=\|Av\|_2\le \|A\|_2.$$ That is, if the $2$-norm is small then any of its eigenvalues is small. But the trace of the matrix is the sum of the eigenvalues, which is small, because any of the ...


1

Given an $m \times n$ matrix $A$ (with $\text{rank}(A) \ge n$), and an $m \times 1$ vector $b$, the $n \times 1$ vector $x$ which minimizes $\|Ax-b\|_2^2$ is given by $\hat{x} = (A^TA)^{-1}A^Tb$. If there is an exact solution to $Ax = b$, then $\hat{x} = (A^TA)^{-1}A^Tb$ will give you that solution. As far as MATLAB implementation, you can simply try "X ...


0

You can use your row reduction skills to solve this. We have 3 equations and 3 unknowns, where the unknowns are $c_1$, $c_2$ and $c_3$. We have: $3c_1+0c_2+0c_3=3,1c_1+4c_2-3c_3=-10,5c_1-3c_2+3c_3=14$. This suggests the augmented matrix: $$ \begin{pmatrix} 3&0&0&3\\1&4&-3&-10\\5&-3&3&14 \end{pmatrix} $$ When row ...


0

Your ad hoc solution is excellent. The general method consists in solving a system of linear equations, reducing it to row echelon form via Gaussian elimination. In your case you start with the matrix $$ \begin{bmatrix} 0&0&3&3\\ 0&3&3&r\\ -3&4&1&-10\\ 3&-3&5&14\\ \end{bmatrix} $$ and apply row elementary ...


0

You are correct, this problem was set up with the first coordinate nice. In general, if you have $n$ vectors in $\Bbb R^n$ with one unknown value, you can just form the determinant and choose the value that makes the determinant zero. This will be a linear equation, but will take a bunch of computation to get there. In your example, for $n=4$, there are ...


2

A vector space is a set with an addition law and a scalar multiplication law, where the scalars are elements of a field. Thus, a vector space over a field may not be itself a field (e.g. continuous functions on an interval); however, a field is always a vector space over itself. Similarly, taking direct sums of a field will give you a vector space over the ...


0

The minimal polynomial of $\begin{bmatrix}0&1\\0&0\end{bmatrix}$ is $z^2$. copper.hat and Daniel Fischer also posted answers in the comments with more generality.


0

Perhaps it is easier to understand this without the $ij$ notation. To show that $M_n(K)$ is the direct sum of $U$ and $W$, we need to show that we can write any matrix $X \in M_n(K)$ as the sum of a matrix in $U$ plus a matrix in $W$ (this is the definition of a direct sum). Note that for any $n \times n$ matrix $X$, we have $X = \dfrac{1}{2}(X+X^T) + ...


3

We'll avoid the metric tensor altogether to keep things simple. The demonstration is simpler this way, although not "proper". The polar coordinate system is rather "nice" to deal with, so you don't need the full machinery of tensors in curvilinear coordinates. First of all, note that W is a scalar. The second fact is that you can think of polar coordinates ...


0

To show that $P_U$ is self adjoint, it's enough to take any matrix representation of $P_U$. For example: if $U$ is a space with basis $u_1,\dots,u_n$, then $$ P_U = u_1u_1^* + \cdots + u_nu_n^* $$ If $P$ is self-adjoint with $P^2 = P$, we can note that all eigenvalues must be $1$ or $0$ and apply the spectral decomposition to find $P = VDV^*$ where $V$ is ...


1

Not a solution but maybe a step in the right direction. First you can formulate your problem as a matrix differential equation. With the matrix vectorization operator, which stacks the columns of the matrix , we write: $$m(x) = \text{vec}(M(x))$$ whereas $m(x)$ is just the vector of your unknown functions $f_i(x)$. Your original problem is $$M'(x) = (A + ...


1

Graphing your vectors on Wolfram Alpha you'll see that you've graphed the second vector incorrectly. Using the image here, you'll see that the angle from vector $2$ to vector $1$ is indeed counter clockwise, and so the Otherwise, if you need for the second vector to point downward, then you should use $[7, -102]$, which will then yield a determinant ...


1

Here's a hint at an alternative proof that $U$ and $W$ are subspaces of $M_n(K)$. Let $T_U,T_W\colon M_n(K)\to M_n(K)$ be given by $T_U(A)=A-A^T$ and similarly $T_W(A)=A+A^T$. Show that these are linear transformations. Show that $\ker T_U=U$ and $\ker T_W=W$. Deduce that $U$ and $W$ are subspaces of $M_n(K)$.


2

Note that $U := \{A \in M_n(K): A^T = A\}$ and $W := \{A \in M_n(K): -A^T = A\}$. So the idea of your friends is the same you had (checking directly that it is closed under multiplication and addition) but you two write it differently. Now, if $K = \mathbb F_2$, note that the matrix $I\in M_n(K)$ with $I_{i,j} = 1$ for every $1 \leq i,j \leq n$, is such ...


3

It will be possible to define multiple norms on any vector space (although they may be equivalent). For example, $||x|| = c\sqrt{\langle x , x \rangle }$ will also be a norm for any $c$. Indeed for your example, there is an infinite class of non-equivalent norms given by $$||P(x)||_p = \left ( \int_0^1\left |P(x)\right|^pdx\right )^\frac1p$$ for any $p \in ...


0

An answer by year! Consider the Riccati equation (1) $XAX+XB+CX+D=0_n$ where $X\in M_n(\mathbb{C})$ is unknown and $A,B,C,D\in M_n(\mathbb{C})$ are generic given matrices. (that is, the entries of these matrices are independent commutative indeterminates OR, more simply, there are no algebraic relations with coefficients in $\mathbb{Q}$ linking the entries ...


0

What does it mean to say a matrix represents a linear transformation??? Suppose $A$ is a matrix for linear transformation $T$ then we should have $Av=Tv$ for all $v\in V$ In case of a $2\times 2$ matrix for a linear transformation on a vector space of dimension $2$ you would have some thing like below.... ...


0

The standard basis for $P_2(\mathbb R)$ (which I assume to be the set of polynomials of degree at most $2$) is $(1,x,x^2)$, and write $e_1=1$, $e_2=x$, $e_3=x^2$. The matrix $M$ of $T$ in the base $B$ has as entry $m_{i,j}$ the coefficient of $T(e_j)$ along $e_i$. For example, $T(e_1) = T(1) = 0 = 0e_1$, so that $m_{1,1} = 0$. Simarly, $T(e_3) = T(x^2) = 2x ...


0

Just take the first two equations and solve that system. There will be up to four solutions (assuming that the two equations are not multiples of one another). Then check if the solutions satisfy the other three equations. To solve $$\begin{align} A_1X^2+B_1Y^2+C_1XY+D_1X+E_1Y+F_1&=0\\ A_2X^2+B_2Y^2+C_2XY+D_2X+E_2Y+F_2&=0 \end{align}$$ let's ...


0

We have two problems here: 1 proving existence of solutions, 2 finding said solutions So begin by subtracting your constant terms first to the left hand side: $$\begin{align} A_1X^2+B_1Y^2+C_1XY+D_1X+E_1Y&=-F_1\\ A_2X^2+B_2Y^2+C_2XY+D_2X+E_2Y&=- F_2\\ A_3X^2+B_3Y^2+C_3XY+D_3X+E_3Y&=-F_3\\ A_4X^2+B_4Y^2+C_4XY+D_4X+E_4Y&=-F_4\\ ...


1

Both are false. For example, consider $m = 3$ and $n = 4$. For part A, consider: $$ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix} \qquad\text{and}\qquad b = \begin{bmatrix} 0 \\ 0 \\ 7 \end{bmatrix} $$ Since the third row of $A$ is all zeroes, any linear combinations of its columns ...


0

Hints. A is false in general, as shown by the system $$\begin{cases} x+y+z=0\\ x+y+z=1 \end{cases}$$ B is definitely false, because you have more unknowns than equations.


1

Hint. Note that linearly (in)depence depends on the scalar field. As an example, lets consider $V = \mathbb C$. Then $V$ is obviously an $1$-dimensional vector space over $\mathbb C$ and $v_1 = 1$ and $v_2 = i$ are linearly depended (for the scalars $\lambda_1 = i$ and $\lambda_2 = -1$ we have $\lambda_1 v_1 + \lambda_2 v_2 = 0$). But over $\mathbb R$ $v_1$ ...


0

I assume that by the phrase "Hermitian symmetric" our OP lovelesswang means what is usually called simply Hermitian, and by the notation $A^\ast$ is meant what is usually denoted by $A^\dagger$, that is, the Hermitian adjoint of $A$, or, in matrix, terms, its conjugate transpose, $A^\dagger = \bar A^T$, where $(\bar A)_{ij} = \bar A_{ij}$. This ...


1

Let $y\in \operatorname{im}T$ so there's $x=x_1 v_1+\cdots+x_n v_n\in V$ such that $$T(x)=T(x_1 v_1+\cdots+x_n v_n)=x_1T(v_1)+\cdots+x_n T(v_n)=y$$ hence we proved that $$\operatorname{im}T\subset \operatorname{span}(T(v_1),\ldots, T(v_n))$$ and the other inclusion is clear hence the equality.


2

Using linear combinations, observe that: \begin{align*} T(b) &= T(-5,2) \\ &= T(\tfrac{1}{2}(-10,3) + \tfrac{-1}{6}(0,-3)) \\ &= T(\tfrac{1}{2}a - \tfrac{1}{6}c) \\ &= \tfrac{1}{2}T(a) - \tfrac{1}{6}T(c) &\text{by the linearity of $T$} \\ &= \tfrac{1}{2}(1,3) - \tfrac{1}{6}(9,3) \\ &= (-1, 1) \end{align*}


0

This isn't an answer but this is how you can solve the problem. The approach that I would do is as follows. We can write a matrix using the vectors that are elements of our subspace as coloumn vectors of the matrix. From what you wrote I believe this is $w_3, w_4, w_6$. So \begin{bmatrix} -2 & -1 & 0 \\ -3 & -1 & 1 \\ -1 & -3 & 1 \\ ...


1

None of the $w_i$ is a multiple of another. So if you can find two of them in the space generated by $v_1$ and $v_2$, you will be finished. This can be attempted like you started. The fact that one of the last two coordinates of the $v_i$ is $0$ makes the job easier. We deal first with $w_1$. If $w_1=av_1+bv_2$, then from the last two coordinates we see ...


1

A combinatoric perspective follows from writing the matrix multiplication in index notation i.e. $(AB)_{ij}=A_{ik}B_{kj}$ (I'll use the Einstein convention, i.e. doubled indices are summed implicitly.) Note that the components $A_{ij}$ of your matrix $A$ equal 1 if $i\leq j$ and zero otherwise. Suppose we now take some arbitrary power of $A$; we will obtain ...


0

After "cross multiplying" and canceling the factor of $\sqrt{2}$ on each side, we have $$ \sqrt{a^2 + b^2 + c^2} = |a - b|. $$ Now, square both sides and remove $a^2$ and $b^2$ terms that appear on both sides, yielding $$ c^2 = -2ab. $$ From the original equation, we know that $a + b + c = 0$, so $$ b = -a - c. $$ At this point, we can make the simplifying ...


0

The plane $x-y=0$ has normal $n_2 =(1,-1,0)$. Suppose the line has unit direction $d$, then $\langle d, n_2 \rangle = {1 \over \sqrt{2}}\|n_2\| = 1$, in particular $d_1 -d_2 = 1$. (There is nothing special about choosing a unit norm direction, but it affects the inner product, so once you choose a length, you must stick with it.) You know that $d$ is ...


7

The question can be rephrased as to show that $\sqrt2$ is not an eigenvalue for $T$. If it were, then there would be a corresponding eigenvector $v$, with $Tv=\sqrt 2\,v$. So for such $v$ you would have $$ \|Tv\|=\|\sqrt2\,v\|=\sqrt2\,\|v\|>\|v\|, $$ contradicting the given inequality. Edit: this answer uses in an essential way the fact that $V$ is ...


3

Another solution is repeated squaring. Strictly speaking this is also repetitive multiplication, but in a much more smart way, so I present it as an answer as well. It only requires 5 matrix multiplications instead of 13. One just has to realize that $A^{13} = A^8 A^4 A$ $A^4$ is given by $(A^2)^2$, so two multiplications and $A^8$ is obtained by one ...


3

Generally in a Banach space, if $\|A\| < 1$, then $I+A$ is invertible (consider $\sum_{k=0}^\infty (-1)^k A^k$). Consider the invertibility of $I - {1 \over \sqrt{2}} T$ (what $A$ matrix, and why does it have suitable operator norm?) Explicitly: We have $R = T-\sqrt{2} I = -\sqrt{2}(I-{1 \over \sqrt{2}} T) $. Let $S = - {1 \over \sqrt{2}} ...


0

Here is a different approach that gives more information. It shows that the algebraic multiplicities match up as well. I'll defer to previous answers for the fact that a matrix is invertible iff $0$ is not an eigenvalue, which will be used implicitly below. I just focus on the eigenvalues of the inverse. Suppose $A$ is an invertible $n \times n$ matrix. ...


1

Since the thing you are minimizing is an everywhere-positive quadratic function of $t$ and $s$, it is convex in each variable. So, we need its critical point, and failing that, something close to it. The function has a critical point when the vector connecting points on two lines is orthogonal to each line. At this point, the vector $$v = P1 + s(P2-P1) - ...


4

For one thing, that limit isn't typically going to exist, even when the function is differentiable. Take $$g(x,y) = x \, .$$ Consider $h^1_n = (\frac{1}{n},0)$, $h^2_n = (0,\frac{1}{n})$. If you take the limit along the sequence $(h^1_n)_{n=1}^\infty$, the limit is $1$. Taking it along $(-h^1_n)_{n=1}^\infty$, the limit is $-1$. Taking it along ...


0

Tu prueba es correcta. Una forma más rápida de concluir es recordando que para cualquier espacio vectorial $V$, si $U$ es un subespacio de $V$ entonces $\dim U \leq \dim V$, y, en el contexto de tu pregunta, $\operatorname{Im}(T)$ es siempre un subespacio de $W$. Your proof is correct. A quicker way to finish the argument is by recalling that for any ...


0

The following results are in this paper: Let $S$ be any set of commuting matrices. Then there is a fixed nonsingular matrix $P$ such that $P^{-1}AP$ is upper triangular for each $A$ in $S$. Let $A_j$, $1\le j\le p$, be commuting $n\times n$ matrices, and let $f=f(x_1,\dots,x_p)$ be an arbitrary polynomial in $x_1,\dots,x_p$. Then there is a fixed ...


9

Although not explicitly said in the question, we are assuming that $V$ is an inner product space. Suppose that $v_1,\ldots,v_n$ is linearly dependent. Then there exist constant, not all zero, $c_1,\ldots, c_n$ with $c_1v_1+\cdots+c_nv_n=0$. Then $$ ...


1

This, to be sure, is one consequence of the rank nullity theorem. However, assuming you haven't learned that yet, consider the following: Let $\{v_1,\dots,v_n\}$ be a basis of $V$. Then $\{T(v_1),\dots,T(v_n)\}$ must span the image of $T$, which has dimension at most $\dim(W)$. So, the vectors $\{T(v_1),\dots,T(v_n)\}$ can't be linearly independent. ...


1

$F$ has dimension $2$ because it's the $0$-set of a linear form and $G$ has dimension $1$ Since $g$ does not satisfy the equation $x-y+z=0$, $g\notin F$ so that $F\cap G=\{0\}$ and the sum $F+G$ is direct. Finally $\Bbb R^3=F\oplus G$ because $E=\Bbb R^3$ and $F\oplus G$ have both dimension $3$.


0

You're right. Remember that $$ \text{rank } T + \text{nullity } T = \text{dim } V. $$ What is the maximum possible rank of $T$? (It may help to think of $T$ as an $m \times n$ matrix.)


0

Hint: suppose that $w_{n-1}$ is such that $\langle w, v_j \rangle > 0$ for $j = 1,\dots,n-1$. Let $S_{n-1}$ be the span of $v_1,\dots,v_{n-1}$. Suppose furthermore (without loss of generality) that $w \in S_{n-1}$. Let $v^\perp$ be the component of $v_n$ perpendicular to $S_{n-1}$. Let $w_{n} = w_{n-1} + a v^\perp$ for some constant $a>0$. We may ...


3

The matrix is the sum of the identity and a nilpotent upper triangular matrix, $I + N$ where: $$ N = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} $$ Since $(I+N)^{13} = I^{13} + 13 I^{12} N + \binom{13}{2} I^{11} N^2 + \binom{13}{3} I^{10}N^3 + \ldots$, it suffices to note: $$ N^2 = \begin{pmatrix} 0 & 0 ...



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