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7

As others have noted, since $A=(a_{ij})_{i,j=1\dots n}$ has determinant $$ \det A = \sum_{\sigma\in S_n} \epsilon_\sigma\prod_{k=1}^n a_{k,\sigma(k)} $$ which is a polynomial expression in the $a_{ij}$, the map $\det: \mathbb R^{n\times n}\to\mathbb R$ is infinitely differentiable. The first derivative with respect to $a_{ij}$ is calculated as $$ ...


3

Yes, diagonalize the zero matrix. THM Let $A$ be a matrix associated to a linear transformation $T:V\to V$, $V$ a $k$-vector space. Let $\lambda_1,\ldots,\lambda_l$ be it's distinct eigenvalues. Then $A$ is diagonalizable iff $$V=E(\lambda_1)\oplus\cdots \oplus E(\lambda_l)$$ where $E(\lambda_i)=\ker(A-\lambda_i1)$. The sum is always direct, but it ...


3

A matrix is singular if and only if $0$ is one of its eigenvalues. A singular matrix can be either diagonalizable or not diagonalizable. For example, $$ \pmatrix{ 1&0\\0&0 } $$ Is diagonalizable (since it is diagonal), whereas $$ \pmatrix{ 0&1\\0&0 } $$ is not diagonalizable.


3

What about this linear transformation: $$T(x,y)=(x,y,-x-y)\quad?$$ Added The set $\{(x_1,x_2,x_3)\in \mathbb{R}^3:x_1 + x_2 + x_3 = 0\}$ is a subspace of $\Bbb R^3$ since it's the kernel of the linear form $$(x,y,z)\mapsto x+y+z$$ hence this set is a hyperplan of $\Bbb R^3$ and its dimension is $2$.


2

We know that $v_1,v_2,v_3$ are linearly independent. This means that $x_1v_1+x_2v_2+x_3v_3=0$ only has the solution $x_1=x_2=x_3=0$. In particular, if we put $x_3=0$, then $x_1v_1+x_2v_2+0v_3=0$ would have also only the solution $x_1=x_2=0$. This proves that $v_1,v_2$ are linearly independent. For any other subset, we can put the zero in the other $v_i$ ...


2

Your RREF tells you that $S$ is linearly independent. What you have done is row-reduced the augmented matrix: $$\left(\begin{array}{ccc|c} 1 & 4 & -2 & 0\\ 3 & 1 & 5 & 0\\ 0 & 2 & -2 & 0 \end{array}\right)$$ to the reduced row echelon form $$\left(\begin{array}{ccc|c} 1 & 0 & 2 & 0\\ 0 & 1 & -1 & ...


2

For the first part of the question, note that there is an isomorphism between $\mathcal{L}(V,V)$ and $\mathbb{F}_{n \times n}$ (matrices of order $n \times n$ over $\mathbb{F}$), which we can denote by $\Phi$ sending each $\tau \in \mathcal{L}(V,V)$ to its matrix representation with respect to some basis $\beta$ of $V$. So dim$(\mathcal{L}(V,V))=$ ...


2

If you would not know that the determinant is a polynomial in the entries of the matrix you may know that it is, if considered as a function of the columns (or rows) of the matrix, mulitilinear, hence $C^{\infty}$ as a function of the columns. Since the matrices depend smoothly on their entries they also depend smoothly on the columns.


2

For the first eigenvalue, $\lambda_1 = 4$, we form $[A-4I]v_1=0$ and we have a RREF of: $$\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right)v_1 = 0 \implies v_1 = (1,0)$$ For the second eigenvalue, $\lambda_2 = 1$, we form $[A-1I]v_2=0$ and we have a RREF of: $$\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ \end{array} ...


2

Call your $(n+1)\times(n+1)$ matrix $B$. If $B$ is singular, then its last column is a linear combination of the first $n$ columns, i.e. $Au=x$ and $x^Tu=0$ for some vector $u$. Thus $x^TA^{-1}x=x^Tu=0$, which is impossible because $A$ is positive definite and $x$ is nonzero. Therefore $\det(B)\ne0$. If $\det(B)>0$, then all leading principal minors of ...


1

We can use Doolittle's Method: $$\begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} \cdot \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix} = \begin{bmatrix} 3 & -7 & -2 \\ -3 & 5 & 1 \\ 6 ...


1

Since $\operatorname{rank}(T)=1$ then by the rank-nullity theorem we have $$\dim\ker(T)=n-1$$ and since $$\operatorname{Im}(T)\cap \ker(T)=\{0\}$$ then $$V=\operatorname{Im}(T)\oplus \ker(T)$$ now we take a basis of $V$ adapted to the previous decomposition and we see that the matrix of $T$ in this basis take the form ...


1

This is unfortunately not a "complete" answer, but there will not be enough space for me to do this in the comments, and at least, it is a partial answer...first, let me just lay down the boundaries here: I will provide an outline for a proof that a unit k-dimensional hypercube will always fit into an ORTHOGONAL projection of a n-dimensional hypercube onto a ...


1

Suposse that 0 is an eigenvalue of A. Considering $J$ the Jordan canonical form of A, you have a triangular matrix with the eigenvalues in the diagonal, with the same determinant than $A$. It's known that determinant of a triangular matrix is the product of the elements in the diagonal and we've supossed that one of them is 0, then is clear that ...


1

We are given the nonhomogeneous system: $$x'(t)=\begin{bmatrix} 2 & -5\\1 & -2 \end{bmatrix}\vec{x} + \begin{bmatrix} \csc t\\ \sec t \end{bmatrix}$$ We can write this as $X'(t) = A x(t) + F[t]$, where: $$A = \begin{bmatrix} 2 & -5\\1 & -2 \end{bmatrix}, ~~ F[t] = \begin{bmatrix} \csc t\\ \sec t \end{bmatrix}$$ We first find the ...


1

That polynomial is called the minimal polynomial. Notice that $1$ and $3$ are the eigenvalues. They are $2$ and they are different. Therefore the characteristic polynomial is the minimal polynomial. $f(x)=(x-1)(x-3)$. Alternative not knowing much. We need $f(x)=x^n+a_{n-1}x^{n-1}+...a_1x+a_0$ such that $f(A)=0$. An equation like $f(A)=0$ will give us ...


1

The matrix of $T$ with respect to the basis $b_i$ consists of the columns $[6 \  7]^T$ and $[3 \ 8]^T$. If $B$ denotes the matrix transforming a vector given in the standard basis into a vector in the basis $b_i$ then the matrix of $T$ in the standard basis will be $$ B^{-1}TB$$ It is not difficult to find $B$. I would like to leave it to you.


1

From what you've written, the matrix of $T$ with respect to the basis $B=\{b_1,b_2\}$ should be $$ [T]_B = \begin{bmatrix} 6 & 3 \\ 7 & 8 \end{bmatrix}. $$ So to write the matrix with respect to the standard basis, it would be helpful to find the change of basis matrix, call it $P$, taking a matrix written with respect to $B$ to one written with ...


1

The key flaw in your argument is that (bi)linearity isn't preserved under duplication of arguments; if $f(x,y)$ is a bilinear function of $x$ and $y$ then that doesn't imply that $g(x)=f(x,x)$ is a linear function of $x$. The easiest way to see this is to specialize to the one-dimensional case and look at $f(x,y)=xy$; this is obviously bilinear in $x$ and ...


1

Both expressions are quadratic on $x$, not linear. So if $x=\sum x_je_j$ where $e_j$ are the canonical basis vectors, you have $$ x^TAx=\sum_j\sum_kx_jx_k\,e_j^TAe_k. $$ Similarly, $$ \text{Tr}(Axx^T)=\sum_j\sum_kx_jx_k\,\text{Tr}(Ae_ke_j^T). $$ Now you could verify that $e_j^TAe_k=\text{Tr}(Ae_ke_j^T)$ to finish the proof. This is certainly easy, but I ...


1

This is an amplification of Lemur's answer, expanding on the "see your vectors and play with them" in his comment. We want to know if we can find scalars $\alpha,\beta,\gamma$ such that $$\alpha(1,3,0)+\beta(4,1,2)+\gamma(-2,5,-2)=(0,0,0)\ .$$ But looking at the third component gives immediately $\beta=\gamma$. So the previous equation can be written as ...


1

Observe that such matrices $A$ satisfy $A^2 = 0$; this implies any eigenvalue $\mu$ of $A$ must vanish, since $Av = \mu v$ for $v \ne 0$ implies $0 = A^2v = \mu^2 v$, forcing $\mu = 0$ since $v \ne 0$. But then if $D = PAP^{-1}$ were diagonal, we would have $D =0$, since the diagonal entries of $D$ must all be eigenvalues of $A$. And this follows from the ...


1

A canonical bases is something like a "natural/obvious choice" of a basis. So you cannot prove, that something is a canonical bases, you can just state, that most of the mathematicians would use take a given basis as a canonical one. Your choice for the canonical bases are right. Note, that if you take a transformation $f$ which maps ...


1

Linear dependence does not say that all vectors in the set can be written as a non-trivial linear combination of the other vectors in the set. Consider $\{ (1, 0, 0), (2, 0, 0), (0, 1, 0) \}$. Notice the first two vectors are linearly dependent, which implies the set is linearly dependent. However, $(0, 1, 0)$ is independent of the other two vectors. So if ...


1

A set of vectors is linearly dependent if and only if there is a nontrivial linear combination of the vectors in the set which is zero. Nontrivial here means that there is at least one nonzero coefficient in the linear combination. It should be immediately obvious then that you can append ANY set of vectors, finite or infinite, even uncountable, to a ...



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