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0

The skew symmetric metric is a matrix with entries like each ij th entry is equal to the (-ji )th entry . We get possible no.of base elements =(n-1)+(n-2)+(n-3)+........+[n-(n-1)] =(n-1)+(n-2)+(n-3)+.........+1 =sum of integers from 1 to n-1 numbers ...


1

I think you've mis-stated some of your conditions. As I understand them, the following two matrices satisfy your three conditions but they are not similar: $$\begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \qquad \text{and} \qquad \begin{bmatrix} 2 & 2 \\ 1 & 2 \end{bmatrix}$$


1

Generally, $\operatorname{tr} (AB) = \sum_k [AB]_{kk} = \sum_k \sum_i [A]_{ki}[B]_{ik} = \sum_i \sum_k [A]_{ki}[B]_{ik} = \sum_i \sum_k [B]_{ik} [A]_{ki} = \operatorname{tr} (BA)$.


1

The original system of equations is equivalent to $\begin{cases}\alpha+\beta=a-c\\\alpha+\beta=b\\\phi=c\end{cases}\iff \begin{cases}\alpha+\beta=b\\b+c=a\\\phi=c\end{cases}\iff \begin{cases}\alpha=\beta-b\\b+c=a\\\phi=c\end{cases}$. The only equation where $\beta$ acts is in the expression of $\alpha$. Therefore, $\alpha$ must depend on $\beta$ and you ...


0

Hint Let denote $B^*_c=(e^*_1,e^*_2,e^*_3)$ the dual basis of $(1,x,x^2)$ then we have $$b_1=\alpha_1e^*_1+\beta_1e^*_2+\gamma_1e^*_3$$ hence $$b_1(1)=1=\alpha_1\quad;\quad b_1(x)=0=\beta_1\quad;\quad b_1(x^2)=0=\gamma$$ and do the same thing for the other vectors and you get the column vectors for the various $[b_i]_{E^∗}$.


0

If you see that knowing that $\text{Im}(A+B)\subseteq \text{Im}(A)+\text{Im}(B)$ is sufficient, then notice that, for any vector $x$, we'd have that $(A+B)x=Ax+Bx$. Note that $Ax$ is in $\text{Im}(A)$ and $Bx$ is in $\text{Im}(B)$. Their sum is therefore in $\text{Im}(A)+\text{Im}(B)$, implying that $\text{Im}(A+B)\subseteq \text{Im}(A)+\text{Im}(B)$.


1

Note that $(T(A))^t=T(A^t)=A^t+2A$. Then solve for $A$ from these two equations in terms of $T(A)$ and $T(A^t)$...


1

Question 1 is just computations. For question 2 you want to show that $T$ is injective, which will suffice; so suppose $T(A)=0$, or $$ A=-2A^T $$ Transposing we get $A^T=-2A$, so from the other equation we have $A=4A$ that implies $A=0$. The same idea can be used for finding the inverse: you want a solution for $B=A+2A^T$. Transposing we have $$ ...


0

$$\text{tr}(AA') = \sum_{i=1}^{i=n}(AA')_{ii}=\sum_{i=1}^{i=n}\sum_{j=1}^{j=m}A_{ij}A'_{ji}=\sum_{i=1}^{i=n}\sum_{j=1}^{j=m}A_{ij}^2$$ Similarly, $$\text{tr}(A'A) = \sum_{i=1}^{i=n}\sum_{j=1}^{j=m}A_{ij}^2$$


1

Let $A=(a_{ij})$ and $AA'=(c_{ij})$ then $$c_{ij}=\sum_{k=1}^na_{ik}a_{jk}$$ $$\operatorname{tr}(AA')=\sum_{i=1}^n c_{ii}=\sum_{i=1}^n\sum_{k=1}^n a^2_{ik}=\operatorname{tr}(A'A)$$


1

$T(x)=Ax$ is a linear transformation so if $u,v$ are solutions to $Ax=b$, then so is $1/2(u+v)$, as $T(1/2(u+v))=1/2(Au+Av)=1/2(b+b)=b$


0

I'm not sure the question is correct. Testing with xavierm02's suggested example, let $X = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$. Then $$ f(a_1, a_2) = \begin{pmatrix} \bar a_1 & \bar a_2 \end{pmatrix}\begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix} \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} = \bar a_1 i a_2 - \bar a_2 i a_1 $$ So ...


0

Have you studied abstract algebra? I believe it is a natural step after learning linear algebra, as a vector space is often one of the (if not the) first abstract algebraic structure a student encounters. Perhaps take a look at Dummit and Foote's text. It also covers much of what an "advanced" linear algebra text might, and much more.


0

My justification would be as follows. Given this: $$\sum_k \left (\sum_l x[l]h_1[n-k-l]\right)h_2[k]$$ Notice what the evaluations of $x$ and $h_2$ each depend on. $x$ depends exclusively on $l$, and $h_2$ depends exclusively on $k$, respectively. Thus, with respect to the other indexer ($k$ for $l$, and vice versa), the evaluation is a constant (i.e. ...


0

If sums go in a row then you can change summation without any problems. $$ \sum_ng(n)\sum_mf(m)=\sum_n\sum_mf(m)g(n) = \sum_m\sum_nf(m)g(n) = \sum_mf(m)\sum_ng(n) $$ If you want a strict proof then you have to count summands in left and right sides. But in this case it is obvious. You are summarizing over a square grid and there is no difference to sum it ...


1

Tensors have "ranks". A rank 0 tensor is just a number, a rank 1 tensor is a vector, a rank 2 tensor is a matrix (vector of vectors), a rank 3 tensor is a vector of vectors of vectors, and so on.


4

Please refer to the trigonometric inverse functions, especially arcsine and arccosine. Given that $\cos t = c_1$ and $\sin t = c_2$, Fundamentally, for a certain range, $\displaystyle t = \arccos(c_1) = \arcsin(c_2)$ Since $\sin(x)$ and $\cos(x)$ are many-to-one functions, there will correspond multiple values of $x$ that yield a certain $\sin(x)$ or ...


1

Here's one sequence of ideas. In single variable calculus we approximate a function locally by a linear function (or, technically, an affine function). To extend this idea to a multivariable setting, we must introduce the idea of a linear transformation. Then, matrices arise as a concise way to describe linear transformations. The determinant can be ...


1

One can measure a signal $y$ (for example the temperature; or the price of a stock at the stock market) discretely as a function of time $t$, with a fixed sampling period $\delta t$. This way one obtains a set of $N$ measured values $y(n)$, $n= 1,2, 3,... N$. These measured values will exhibit behaviour at different time scales. For example the outside ...


1

You can have $m=n+1$. Take the vertices of a regular simplex with centre at the origin. You can't have $m=n+2$. There is at least a two-dimensional space of vectors $(a_1,\ldots,a_{n+2})$ such that $$\sum_{i=1}^{n+2} a_i v_i=0.$$ This gives enough room for manoeuvre to ensure some $a_i>0$ and some $a_j<0$. Thus we get some nontrivial relation ...


0

Affine Geometry works well to smoothly fit curves and surfaces to a set of desired data points. Affine Geometry builds on top of Linear Algebra and is essential to any engineering that combines absolute and relative for both direction and displacement. Engineering the moving parts of a vehicle combine absolute and relative motion with both rotational and ...


0

Note that $x'=0$ when $x^2+y^2-2=0$, i.e., on the circle $x^2+y^2=2$. So outside the circle, $x'>0$ and inside the circle, $x'<0$. This tells you the "east-west" behavior of trajectories in those regions. On the other hand, $y'=0$ on $y=x^2$ so above the parabola, $y'>0$ and below it $y'<0$. This tells you the "north-south" behavior of ...


2

Let $\{v_1, v_2, \cdots , v_n \}$ be a set of linearly dependent vectors of a vector space $V$ with $v_i \neq 0, \forall i = 1, 2, \cdots , n$. Then by Linear Dependence Lemma there exists a vector $v_i$ which can be written as a linear combination of rest of the $v_j$'s. But we can also choose a $v_i$ with the property that it can be written as a linear ...


0

Hint. Consider any $2\times2$ rotation matrix $A=\pmatrix{\cos t&-\sin t\\ \sin t&\cos t}$ for an angle $t$ such that $\sin t\ne0$.


1

This is not true in general. Consider for example the family of real skew-symmetric matrices i.e. $A=-A^T$. Clearly skew-symmetric matrices satisfy $AA^T=A^TA$. The given matrix $B$ for this family is: $$B=2AA^T(A^T-A)=-2A^2(-2A)=4A^3.$$ While the matrix $B^T$ for this family is: $$B^T=2(A-A^T)AA^T=2(2A)(-A^2)=-4A^3.$$


2

It sounds like a magic matrix is a special case of a doubly stochastic matrix, (and in turn a special case of a stochastic matrix), scaled by a scalar value: the magic constant. A stochastic matrix is just a square matrix of nonnegative real numbers with each row summing to $1$. The Perron–Frobenius theorem ensures that every stochastic matrix has an ...


4

We have $$\theta = \sum_{i=1}^{n-1} x_i \otimes x_{i+1} - \sum_{i=1}^{n-1} x_{i+1} \otimes x_i$$so$$\theta(A, B) = \sum_{i=1}^{n-1} A_iB_{i+1} - \sum_{i=1}^{n-1} A_{i+1}B_i = \sum_{i=1}^{n-1}(-1)^{i+1} - \sum_{i=1}^{n-1}(-1)^i = \left\{ \begin{array}{ll} 2 & \mbox{if } n \text{ even}\\ 0 & \mbox{if } n \text{ odd} \end{array} \right..$$


0

Let's write it out fully: $B=2AA^tA^t-2AA^tA$. At this point use $AA^t=A^tA$ to get: $B=2A(A^t)^2-2(A^2)A^t$. So, $B^t=2A^2A^t-2A(A^t)^2=-B$.


0

$$ \begin{eqnarray*} B^T&{}={}&2\left(A-A^T\right)AA^T\newline &{}={}&2\left(AAA^T-A^TAA^T\right)\newline &{}={}&2A\left(AA^T-A^TA^T\right)\newline &{}={}&2AA^T\left(A-A^T\right)\newline &{}\neq{}& B\,. \end{eqnarray*} $$ Almost, but not quite, in general.


0

Let $A$ be a $(n\times n)$ matrix that satisfies: $AA^t=A^tA$ Let $B$ be a matrix such that: $B=2AA^t(A^t-A)$ Prove/disprove that: $B^t=B$ HINT: $$\begin{align} B &=2AA^t(A^t-A) \\ B^t &=((2AA^t)(A^t-A))^t \\ B^t &= (A^t - A)^t(2AA^t)^t \\ B^t &= 2(A - A^t)A^tA \\ B^t &= 2(A - A^t)AA^t \\ B^t + B &= 2((AAA^t - A^tAA^t - ...


0

Continuing from where you left off: $B^t = AAA^t2 - A^tAA^t2$ $B^t = 2AAA^t - 2A^tAA^t$ ($2$ is a scalar, so it commutes with any matrix) $B^t = 2AA^tA - 2AA^tA^t$ (Use the fact that $AA^t = A^tA$) $B^t = 2AA^t(A - A^t)$ (Factor the expression) $B^t = -2AA^t(A^t - A) = -B$ So we get $B^t = -B$. Thus, the claim $B^t = B$ is true if and only if $B = ...


0

The statement is that the polynomials $\phi(A, \lambda)$ and $\phi(B, \lambda)$ are equal. Two polynomials are equal if and only if all of their coefficients are equal. As $\phi(A, \lambda)$ and $\phi(B, \lambda)$ are polynomials in $\lambda$, they both have a coefficient of $\lambda^{n-1}$. As $\phi(A, \lambda)$ and $\phi(B, \lambda)$ are equal, these ...


0

Compute the transpose of $B$: $$ B^T = (2AA^T(A^T-A))^T = (A^T-A)^T(2AA^T)^T = (A-A^T)2A^TA = 2(AA^TA - A^TA^T A) = 2(AA^TA - AA^TA^T) = 2AA^T(A-A^T).$$ Then $$B-B^T = 4AA^T(A^T-A).$$ So the claim is true if and only if $A^T=A$, i.e. $A$ is symmetric.


1

One way to easily see the first two derivatives of a vector or matrix functional, particularly of a quadratic form, is to use a variational approach. In this case we have $$f(a+\delta a)=(a+\delta a)^HX(a+\delta a)=a^HXa+(\delta a)^HXa+a^HX(\delta a)+(\delta a)^HX\delta a$$ The linear term gives us the gradient: $$\langle \nabla f(a), \delta a \rangle = ...


1

@ EKH , your derivative is false. There are $3$ mistakes: 1. $Xa^*$ is not defined. 2. When you derive $a^2$, there is a factor $2$... 3. Here the result is real. Let $\phi:a\rightarrow a^*Xa$. Then $D\phi_a:h\rightarrow h^*Xa+a^*Xh=2Re(a^*Xh)$. Thus $\nabla_{\phi}(a)$ is associated to $2Re(Xa)$. Let $X=Y+iZ$ where $Y,Z$ are real and $Y=Y^T,Z=-Z^T$, ...


0

I really like Introduction to Matrix Analysis by Richard Bellman.


2

Since we're considering $S \subset \Bbb M_2(\Bbb R)$, it is natural to assume that the operations on $S$ are inherited from $\Bbb M_2(\Bbb R)$. But $S$ as it is, is not a subspace of $\Bbb M_2(\Bbb R)$. We have that $$\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1\end{bmatrix}+\begin{bmatrix} a_2 & b_2 \\ c_2 & d_2\end{bmatrix} = \begin{bmatrix} ...


10

Theorem: every vector space has a basis (and any two bases of the vector space have the same cardinality). This theorem is the cornerstone of linear algebra, and usually the case of finite-dimensional vector spaces is proven in a linear algebra class. (I should warn that the above isn't true in general unless you assume the axiom of choice.) Your specific ...


2

I would go for some later chapters of Advanced Linear Algebra, from Steven Roman, and Linear Algebra and Geometry, from Kostrikin. About open problems in Linear Algebra, you can take a look at the comments in this question: Are there open problems in Linear Algebra?. Particularly, I find it difficult to find open problems in linear algebra, since (in my ...


0

I am going to give a hint for part 1 to get you started. Based on the approach you followed, I am not sure whether you have solved that part. Note, for example that the fact that all entries of $u$ are nonzero, combined with the fact that $q$ is a nonzero vector does not automatically imply that $u^{T}q \neq 0$; the two vectors could have positive and ...


3

Yes, applying some differential operators to a real-value function, one can get a complex-valued result. This is because the differential operator may happen to have complex coefficients. Let's stick to 1-dimensional case for simplicity. Writing $a=x+iy$, we have $$ \frac{\partial f}{\partial a} = \frac12 \frac{\partial f}{\partial x} ...


2

Let's make a more detailed analysis of what we mean by vector, matrix and tensor. To keep things simple, let's say we are working with vector spaces over $\mathbb R$, and with finite-dimensional vector spaces. A vector is an element of a vector space. Note that a vector space $V$ does not need to be "a tuple $(x_1,x_2,\dots,x_n)$ where $x_i \in \mathbb ...


3

Hint: The matrix is a Vandermonde matrix. So: $$\begin{vmatrix} 1&x&x^{2} \\ 1&y&y^{2} \\ 1&z&z^{2} \end{vmatrix} = \begin{vmatrix} 1& 1 & 1 \\ x & y & z \\ x ^2 & y^2 & z^2\end{vmatrix} = (y-x)(z-x)(z-y).$$ Now analyzing $\det A = 0$ is easy. Look at the intersection of the sphere with the planes $x = y$, $z ...


1

Formally, a tensor is a multilinear, real-valued function of vectors; this, I find, is the best guide to intuition. For it means that once you can identify some collection of objects as having the properties of a vector space, then, to find a tensor, all you need to do is identify linear operations that can be performed on these vectors, where these ...


1

Multiplying the first row by $s$ and subtracting it form the second row...


2

Just multiply the first row by $-s$ and add it to the second row, and you get the matrix $B$. The row-operation is: $R_2\mapsto R_2 - s \cdot R_1$


1

This is the Cauchy-Schwarz Inequality (with a square root taken on both sides).


5

Frobenius Norm Case We first introduce a lemma. Lemma 1. If $A, B$ are respectively $m \times n$ and $n \times p$ matrices, then $\|AB\| \le \|A\|\|B\|$. Proof. Note that $AB$ has columns $A\vec{\beta}_1, \dots, A\vec{\beta}_p$ where $\vec{\beta}_j$ is the $j$th column of $\beta$. So$$\|AB\|^2 = \sum_{j=1}^p \|A\vec{\beta}_j\|^2 \le \sum_{j=1}^p ...


2

Alas, no. Consider the sequence $$ -1, 1, -1, t_0 = 1, t_1 = -1, 1, ... $$ in which you compute a moving average over an interval of size 2, such as $(-1, 1]$. The moving average function will be constantly 0. Since you can multiply the original sequence by any number $K$, you can make the sequence arbitrarily volatile and still have a moving average that's ...


1

The key is to note that, by definition: $$ \| x\|^2= \langle x,x\rangle $$



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