# Tag Info

6

For any two specific vectors, no $C$ does NOT have to be the identity matrix. For example if one of the vectors is $0$ then $C$ can be anything. The only way to talk about $b^{-1}$ is if $b$ is one dimensional and not $0$.

4

Because the range of $T$ contains the range of $T^2$, what you get is that the range of $T$ and of $T^2$ coincide. Because of the rank nullity formula, you also know the kernel of $T$ and $T^2$ coincide. Pick an arbitrary vector $v\in V$. Because $Tv$ is in the image of $T$, there is some $w$ such that $Tv=T^2w$. This reads $T(v-Tw)=0$. It follows that you ...

4

Yes, that's a great way to think about it. The matrix is like the minimal information that you need to write down in order to fully specify the linear transformation.

4

Let us do it over the real numbers - it has been noted in a comment that this does not hold in characteristic 2. Let $v$ be any vector of the underlying vector space $V$. Then $$\tag{eq} v = \dfrac12 \left((A - I) v - (A + I) v\right).$$ (BTW, this does not hold when you cannot divide by $2$, i.e., in characteristic $2$.) Moreover each vector $(A - I)v$ ...

4

A question I am fond of asking my students: Does the vector space of polynomials of degree $\leq n$ have a basis consisting of polynomials with same degree? Answer: yes.

4

First question, let $\{v_i\}$ be a basis, then $\vec{a}=\sum_i a_i v_i$ and $\vec{b}=\sum_i b_i v_i$. Let $C_{ij}=v_i\cdot v_j$, this is for example $\delta_{ij}$ if the basis is orthonormal, but we're not assuming that. Then by bilinearity of the dot product: $\vec{a}\cdot \vec{b}=\sum_i\sum_j a_ib_j v_i\cdot v_j=\sum_i\sum_j a_ib_j C_{ij}$. The $C_{ij}$...

3

Think of the given matrix as a block matrix $$\left[\begin{array}{ccccc|c} b_1 & b_2 & b_3 & \cdots & b_{n-1} & 0 \\ \hline a_1 & 0 & 0 & \cdots & 0 & b_1 \\ 0 & a_2 & 0 & \cdots & 0 & b_2\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ ... 2 b_1 \det\begin{pmatrix} 0 & 0 & \cdots & 0 & b_1 \\ a_2 & 0 & \cdots & 0 & b_2\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n-1} & b_{n-1} \\ \end{pmatrix} - a_1 \det\begin{pmatrix} b_2 & b_3 & \cdots & b_{n-1} &... 2 I'll give you a bunch of hints to start: When computing the determinant, it is best to start with the row containing the greatest number of null entries. In your case it's the last line. First step yields:$$\det(A(n))=b_{n-1}\left|\begin{matrix} b_1 & b_2 & b_3 & \cdots & b_{n-1} \\ a_1 & 0 & 0 & \cdots & 0 ...

2

Don't know how difficult these need to be. Here are some rather simple ones: For every linear map $f : \mathbb R^m \to \mathbb R^n$ there is a unique $m\times n$ matrix with $f(v) = Av$ for all $v\in \mathbb R^m$. Answer: No, its actually an $n\times m$ matrix. The expression $Av$ would not even make sense. If $AB = AC$ and $A\neq 0$, then $B = C$ for all $... 2 No, you are supposed to multiply$A$by itself$k$times using matrix multiplication (in most contexts). So, for example, if $$A = \pmatrix{1&1\\0&1}$$ Then$A^2$will be $$A^2 = AA = \pmatrix{1&2\\0&1}$$ and will not be the same as$A$. 2 There are two ways I would like to think about this problem, which you have essentially highlighted. One is that the function you define is clearly injective and linear, and hence an isomorphism. The other way is that it's really easy to know whether two f.d. vector spaces over the same field are isomorphic; you simply have to check whether they have the ... 2 By the universal property you have a set theoretic map$l : V \times V^* \to V \otimes V^*$,$l(v,f) = v \otimes f$, and a linear map$\phi : V \otimes V^* \to \mathcal{L}(V,V)$such that$\phi \circ l = g$. Hence$\phi$takes$v\otimes f \mapsto vf(\cdot)$. Show$\phi$is an isomorphism. 2 Here is a bit of a more fleshed-out version of what you're saying, from my perspective. What does it mean to take a basis of a finite-dimensional vector space$V$? It means that you identify each vector$v \in V$uniquely with a tuple$(v_{1},\ldots,v_{n})$of components. Now, linear maps work directly with your vector space. Consider the linear map "... 2 Edit It's true in$\Bbb R^n$, by a fairly straightforward argument that actually gives an orthonormal basis. Proof tomorrow... I believe the answer is no. Consider$\Bbb Q^2$, where$\Bbb Q$has the usual order. Say $$P=\{x\in\Bbb Q^2:x_1+\sqrt 2x_2>0\}.$$ Then$P$is closed under addition and multiplication by positive scalars, and for every$x$... 2 The hardest part is finding the point of intersection between the three planes. To do that, you can rewrite it as a matrix equation then apply Gaussian elimination, or you can just do it by hand. Adding the$2$nd and$3$rd equations yields $$5y+7z=19. \tag{1}$$ Adding the$1$st equation and$3$times the$2$nd equation gives $$5y+10z=25. \tag{2}$$ ... 2 For a real-valued and symmetric matrix$A$, then$A$has negative eigenvalues if and only if it is not positive semi-definite. To check whether a matrix is positive-semi-definite you can use Sylvester's criterion which is very easy to check. 2 Usually not. Just consider$\mathbb{C} \oplus \mathbb{C}$; we have lots of projections onto the first component: anything of the form$(x,y) \mapsto (x + cy, 0)$, for$c \in \mathbb{C}$. 1 If your matrix is invertible and positive, then either it already has at least one negative eigenvalue, or you can get a matrix with a negative eigenvalue by exchanging two rows. Proof: If all eigenvalues are positive, then the determinant is positive. Exchanging two rows changes the sign of the determinant. Since the determinant is the product of the ... 1 An$\color{red}{\textrm{ortho}}\color{blue}{\textrm{normal}}$set of vectors is$\color{red}{\textrm{orthogonal}}$and each vector in it is$\color{blue}{\textrm{normalized}}$. Your set has neither of those properties. To orthogonalize your set, use the Gram-Schmidt process. To normalize a vector$\mathbf v$, you divide the vector by its norm:$\frac{\...

1

We have the following formula for the rotation matrix of angle $\theta$ about a vector $u \in \Bbb R^3$. $$R = (\cos \theta) I_3 + (\sin \theta) [u]_{\times} + (1 - \cos \theta) u \otimes u$$ Where $I_3$ is the identity matrix, $[\cdot]_{\times}$ is the cross-product matrix and $\otimes$ is the tensor product. See the Wikipedia page.

1

Use Gaussian Elimination followed by the Euclidean Distance. If you don't know how to use either of these, read up on them from the links provided and from other sources(I recommend Lang's Linear Algebra as a introduction to the subject, but that is slightly off topic). The row echelon form of our system is: $\left(\begin{array}{ccc|c} 1 & \frac{-1}{3} ... 1 In general, let$K \to L$be a field extension.$GL_n(L)/GL_n(K)$can be interpreted as the set of "$K$-structures" on$L^n$. One of many equivalent ways to describe a$K$-structure is that it is a$K$-subspace$V$of$L^n$such that the induced map $$V \otimes_K L \to L^n$$ is an isomorphism. Consequently,$GL_n(K) \backslash GL_n(L) / GL_n(K)$can be ... 1 The column- and row rank of a matrix coincide, so it does not matter whether you transpose the matrix or not. 1 Your question is equivalent to asking whether the dual cone of$A$has a non-empty interior, which is to say that its linear combinations span all of$\Bbb R^n$. The general area you should be looking into is convex optimization. I hope that helps you find some leads. 1 Hints: For a), it is useful to note that$A = xx^T$, where$x$is the column-vector$x = (a,b,c)$. No, the number of minors that are zero is not necessarily the multiplicity of the zero eigenvalue; if you wanted to use minors, you'd have to show that every$2 \times 2$submatrix has determinant zero. You should instead find the rank of the matrix through ... 1 a) I agree. I am not sure about your conjecture how the minors relate to the multiplicity of$\lambda=0$However, I can eyeball two eigenvectors for$\lambda=0v_1 = \begin{bmatrix} b\\-a\\0\end{bmatrix}, v_1 = \begin{bmatrix} c\\0\\-a\end{bmatrix}$b) looks fine c) For each$X$, Suppose$P,Q \in X$then for example iii) if$trace (AP)=0$and$trace (...

1

A simple but not trivial question: (T/F): If $n\times n$ matrix $A$ is of rank $n$, then $A$ is diagonalizable by a similarity transformation $$D = P^{-1}AP$$ Another, slightly harder one: If $P(x)$ is a polynomial and $A$ is a matrix with eigenvalues $\lambda_1 \ldots \lambda_n$, then $P(A)$ has eigenvalues $P(\lambda_1) \ldots P(\lambda_n)$

1

If $T: \mathbb{R}^3 \to \mathbb{R}$ is a linear transformation, then $T(x,y,z) = ax + by + cz$ for some $a,b,c \in \mathbb{R}$. If $S,T: V \to V$ are linear transformations and $v \in \ker T$, then $v \in \ker S \circ T$. If two non-zero vectors are linearly independent, then one must be a scalar multiple of the other. If $T$ is invertible, then $\ker T = ... 1 Let$V$be a vector space over$\Bbb R$. As elaborated in the link in the comment, let$V_{\Bbb C} = V \oplus V$denote the complexification of$V$, in which$v + iw = v \oplus w = (v,w)$. We define the conjugation map by $$J(v + iw) = v - iw$$ For any$v,w \in V$. Note that$J$is$\Bbb R$-linear and that for any$\lambda = a+bi$,$x = v + iw\$, we have: ...

Only top voted, non community-wiki answers of a minimum length are eligible