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5

There are two approaches when taking vector derivatives. First, you can work in coordinates. This will always work, but is not always pleasant. In this case $$S(\beta) = y^Ty - 2\sum_i \beta_i(X^Ty)_i + \sum_{i,j} \beta_i (X^TX)_{ij} \beta_j$$ so \begin{align*} \frac{\partial S}{\partial \beta_k} &= -2\sum_i \delta_{ik}(X^Ty)_i + \sum_{i,j} ...


3

What is meant by the vector derivative $\frac{dF}{d\beta}$ is the vector with components $\frac{dF}{d\beta_k}$. Then $$\frac{d}{d\beta_k}2\beta^T X^T y=\frac{d}{d\beta_k}\sum_{i,j}2\beta_i X_{ji} y_j=\sum_{i,j}2\delta_{ik} X_{ji} y_j=\sum_{j}2 X_{jk} y_j=(2X^T y)_k,$$ so indeed $\frac{d}{d\beta}(2\beta^T X^T y)=2 X^T y$ as desired.


3

Let scalar field $f : \mathbb{R}^n \to \mathbb{R}$ be defined by $$f (x) = a^T x = a_1 x_1 + a_2 x_2 + \cdots + a_n x_n$$ Taking the $n$ partial derivatives, $$\partial_1 f (x) = a_1 \qquad \qquad \partial_2 f (x) = a_2 \qquad \dots \qquad\partial_n f (x) = a_n, \qquad$$ Hence, the gradient of $f$ is $$\nabla f (x) = (a_1, a_2, \dots, a_n) = a$$


3

This is because of the Rank nullity theorem: as matrix and its transpose have the same rank, we have $$\DeclareMathOperator\rk{rank} \rk(A-\lambda I)=\rk{}{}^\mathrm t\mkern-1.5mu(A-\lambda I)\iff \dim \ker(A-\lambda I)=\dim\ker{}{}^\mathrm t\mkern-1.5mu(A-\lambda I).$$


2

To prove: $0=1$. Certain identities get funky when we pass over to infinite-order matrices. We see such matrices, for example, in representations of operators in quantum mechanics. Everyone knows that $Tr(AB)-Tr(BA)=0$. So let $A_{i,j}=\delta_{i,j-1}, B_{i,j}=A_{j,i}$ Here $\delta$ is the Kronecker delta function, and $i$ and $j$ run through all ...


2

Let $ W = \textrm{Im}\, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W \to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ \ker T_W = W \cap \ker T $.


2

Compute: $$\begin{bmatrix} L_1^T \\ L_2^T \end{bmatrix} \begin{bmatrix} X & L_2 \end{bmatrix}$$


2

There can't be a closed form expression here (for any meaning of "closed form" that is weaker than roots of sextic polynomials). For example, try $$ H = \pmatrix{3 & 1 & 0 & 0 & 0 & 1\cr 1 & 3 & 1 & 0 & 0 & 0\cr 0 & 1 & 3 & 1 & 0 & 0\cr 0 & 0 & 1 & 3 ...


2

Let $B$ be an $n\times n$-matrix and let $k:=\dim N(B)$. Then the row-echelon form has all zeroes in its last $k$ rows, so $B^{\top}$ has all zeroes in its last $k$ columns, meaning that $B^{\top}e_i=0$ for the last $k$ basis vectors $e_{n-k+1},\ldots,e_n$. Hence $\dim N(B^{\top})\geq\dim N(B)$ holds for all square matrices $B$. Then $$\dim N(B)\leq\dim ...


2

Is this are you looking for? (using Einstein convention) $$\left[T^a,T^b\right]^i_j=(T^aT^b )_{ij}-(T^bT^a )_{ij}$$ $$=(T^a)^i_p(T^b)^p_j-(T^b)^i_q(T^a)^q_j=\epsilon_{aip}\epsilon_{bpj}-\epsilon_{biq}\epsilon_{aqj}.$$ And then whatever you need to do you probably will need to use the following identity: ...


1

Taken straight from my blog. Recall that the multiple regression linear model is the equation given by $$Y_i = \beta_0 + \sum_{j=1}^{p}X_{ij}\beta_{j} + \epsilon_i\text{, } i = 1, 2, \dots, N\tag{1}$$ where $\epsilon_i$ is a random variable for each $i$. This can be written in matrix form like so. \begin{equation*} \begin{array}{c@{}c@{}c@{}c@{}c@{}c} ...


1

A self-adjoint operator $S : X \to X$ (where $X$ is an inner product space) is an operator such that for all $x,y \in X$, we have $$\langle Sx,y \rangle = \langle x,Sy\rangle.$$ This is a generalization of a real, symmetric matrix. One important property of such operators is that the eigenvalues of a self-adjoint operator are necessarily real. Indeed, if ...


1

The answer hinges on how you define area. One way is to define area for a rectangle, then use approximations by unions of rectangles and take some limiting process to find the area of a more general shape. With this approach, if you know how stretching affects a rectangle, then (skipping all the interesting detail) you can see that it will change the area of ...


1

You can use the Sylvester determinant theorem (or the matrix determinant lemma). It states that $$\det(I+ \beta u u^T J) =\det(1+ \beta u^TJ u) .$$ Now we have that $u^TJ u$ is a $1\times 1$ matrix and thus $$\det(I+ \beta u u^T J) =\det(1+ \beta u^TJ u) = 1+ \beta u^TJ u.$$ Note that so far, we have not used the fact that $J$ is antisymmetric at all. ...


1

It must be a typo. For another reference, if you look at Horn & Johnson's book here (chapter 0, section 0.3.3 in the first edition) the authors discuss how elementary row operations can be achieved via left multiplication by square matrices. Side note: If we use the fact that elementary row operations on a matrix $\boldsymbol{M}$ are equivalent to ...


1

I have my misgivings about your use of the word "proper". But, if what you want is "column-scalar-scalar-row", then you can write $$ B'Axx'A'B. $$ If what you want is an expression $xCx'$, then $C$ may not exist. For starters, $C$ is necessarily $1\times1$, so a number $c$. The equality is then $$ c\,xx'=(Ax)^2B'B=B'AxxA'B. $$ If we multiply by $x'$ on the ...


1

Since $Ax$ is a scalar, so that it equals its own transpose, how about: $$ (Ax)'B'B(Ax)=x'A'B'BAx $$


1

To define a linear map is enough to define it on a basis and expand it linearly. For example the linear map $f:\mathbb R^2\rightarrow \mathbb R^3$ with $f(1,0)=(1,0,0), f(0,1)=(0,1,0)$ is the map $$ f(x,y)=f(x(1,0)+y(0,1))=xf(1,0)+yf(0,1)=x(1,0,0)+y(0,1,0)=(x,y,0). $$ Now define a linear map $g:\mathbb R^3\rightarrow\mathbb R^2$ by specifing what $g(1,0,0), ...


1

(Edit: the OP has modified their question; this answer no longer applies.) Your question is not well posed because determinant is defined on commutative rings only, but $M_n(R)$ in general is not a commutative ring. But there is indeed something similar to what you ask. See John Silvester, Determinants of Block Matrices, theorem 1. Briefly speaking, ...


1

$$e(A)=e(IA)=e(I)A=EA.$$ In fact, in general, you can say $$e(BA)=e(B)A.$$ (This is justified below. The intuition behind this is that applying a row operation to a matrix and then multiplying it with another matrix is the same as applying a row operation to their product.) Alternatively, you can reduce the three cases in the proof you're given to ...


1

(a) Suppose $0$ is an eigenvalue of $A$. Then $Av=0v$ for some nonzero vector $v$. What does that mean? (b) Look at the characteristic polynomial of the matrix $A$ and the characteristic polynomial of its transpose. What can you say about them?


1

Use the implicit function theorem to show that the derivative exists. As for computing it: we will have a very nice derivative if $A'(t)$ commutes with $A(t)$. In particular, we find that $$ \frac d{dt}\sqrt{A(t)} = \frac 12[A(t)]^{-1/2}A'(t) $$ this can be confirmed via the power series for $x \mapsto \sqrt{x}$ centered at $x = 1$, an applying the ...


1

Having $m$ vectors $v_i \in \mathbb{R}^2$ a linear combination that gives a vector $b \in \mathbb{R}^2$ is $$ \lambda_1 v_1 + \dotsb + \lambda_m v_m = b $$ Each linear combination is characterized by the vector $\lambda = (\lambda_1, \dotsc, \lambda_m)^t$. This corresponds with the matrix equation $$ A \lambda=b \quad (*) $$ with $A = (v_1, \dotsc, v_m)$ ...


1

Here is a constructive way to do this. Let $v_1, ... , v_m$ be your vectors, $m > 0$. Let's assume they're all nonzero. Given $c_1, ... , c_m \in \mathbb{R}$, let $$w = c_1v_1 + \cdots + c_mv_m$$ Your question is, what are all the $m$-tuples $r = (r_1, ... , r_m) \in \mathbb{R}^m$ do we have $w = r_1v_1 + \cdots + r_mv_m$? In other words, for which ...


1

Note: You can use the facts that Sum of the eigen values of a matrix is equal to the trace and product of the eigen values of the matrix is equal to the determinant. For the first matrix let $\lambda_1$ and $\lambda_2$ be the eigen values. Then $\lambda_1+\lambda_2=a+d$, the trace of the matrix and $\lambda_1 \lambda_2=ad-bc$, which is the determinant. Now ...


1

Let $z$ be such that $|z^* x| = \|z\|' \|x\| = \|z\|' = \sup_{v:\|v\|=1} |z^* v|$ as given by the lemma. By re-scaling $z$, we may assume $|z^* x|=1$ Then with $B=yz^*$, we have $\|B\| = \sup_{v : \|v\|=1} \|yz^* v\| = \|y\| \sup_{v : \|v\|=1} |z^* v| = \|y\| |z^* x| = 1$. Finally, $Bx = y(z^*x)$. It is not clear to me how we can show that $z^* x=1$; ...



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