# Tag Info

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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