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16

$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}\quad; \quad B=\begin{pmatrix}0&0\\1&0\end{pmatrix}$$ and the generalization is easy: it suffices to take two nilpotent matrices with order of nilpotency $2$.


10

$A = I_2$, $B = \begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}$ Then $A^2 = B^2 = I_2$, but $A \ne \pm B$. Edit I: In general, you can take any two distinct reflection matrices (the ones in Darth Geek's answer, for example). Any such matrix $R$ will have $R^2 = I$ (for they are orthogonal and symmetric). You can find plenty of any order. See for example: ...


8

$$\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix},\; \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}$$


7

Try the vector of all 1's. That'll do it. If this is not clear, then think about it this way. Let $$e_1 = \left( \begin{array}{c}1\\0\\ \vdots \\ 0\end{array} \right), e_2 = \left( \begin{array}{c}0\\1\\ \vdots \\ 0\end{array} \right), ..., e_n = \left( \begin{array}{c}0\\0\\ \vdots \\ 1\end{array}\right)$$ The operation $Ae_1$ is the first column of ...


4

$$\color{red}{\Large\verb/This is a draft/}\\ \small\verb/I haven't verified all details!/ $$ This answer split into 3 parts. The first part review the easy stuff one can do for general $n\times n$ matrices. The second part introduces some extra structure that help us to attack the problem when $C_0$ and $C_1$ are $3\times 3$ skew symmetric matrices. The ...


4

Let $\lambda_{1,2,3}$ denote the eigenvalues of $A$, then $\operatorname{det}(A+I)=\operatorname{det}(A+2I)$ can be rewritten as $$(\lambda_1+1)(\lambda_2+1)(\lambda_3+1)=(\lambda_1+2)(\lambda_2+2)(\lambda_3+2),$$ which is in turn equivalent to (check it!) $$ ...


4

Yes this is correct as long as you assume that $A$ is a $3 \times 3$ matrix, except $tr(A^2) = 1^2+ 2^2 + 4^2 = 21$, since $Av = \lambda v \implies A^2v = \lambda^2v$


3

Try going the opposite direction: Starting with a matrices $A$ and $b$, with components $A_{i j}$ and $b_{i}$ respectively, what would that matrix norm look like when expressed in terms of components? What $A_{i j}$ and $b_i$ will then give you the form you want?


3

What can you deduce about the linear dependence/independence of the columns of $A-sI$?


3

If you compute the determinant by cofactor expansion along the third row, you see that the characteristic polynomial is $$ p_A(\lambda) = |A - \lambda \cdot I| = (1 - \lambda) \cdot \left| \begin{matrix} 1 - \lambda & 1 \\ 1 & 1 - \lambda \\ \end{matrix} \right| = (1 - \lambda) \cdot ((1 - \lambda)^2 - 1) = (1 - \lambda) \cdot (- ...


3

If $a^{2}-b^{2}=0$ in e.g. $\mathbb{R}$ then we have $\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}=0$ and consequently $a=b\vee a=-b$. When it comes to matrices then $\left(A-B\right)\left(A+B\right)=A^{2}+AB-BA-B^{2}$ where $AB$ and $BA$ are not necessarily the same. Secondly $UV=0$ does not necessarily imply that $U=0\vee V=0$ For counterexamples see ...


3

$A = \begin{pmatrix} 1/2 & 1/4 \\ 3 & -1/2\end{pmatrix}$, $B = \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}$


3

Let $a,b\in Sym_{n}$ such that $a^{2}=b^{2}$. Let $A,B$ be permutation matrix corresponding to $a,b$. Then $A^{2}=B^{2}$.


3

If $A^2=I$ then $A$ satisfies $x^2-1$, which means that either $x-1$ divides the minimal polynomial and $A$ has a fixed point (over the rationals, but you can multiply by a constant to get integers) or the minimimal polynomial is $x+1$ and so $A=-I$.


3

The matrix that you looking for is $$A=\left[T(1)\ T(x)\ T(x^2)\right].$$ $T(1)=1+x$ $T(x)=0+x=x$ $T(x^2)=0+x=x$ So $$A=\begin{bmatrix}1&0&0\\1&1&1\\0&0&0\end{bmatrix}.$$ An example is take $p(x)=1+3x^2$, then $$T(p(x))=p(0)+p(1)x=1+4x.$$ By other hand, ...


2

Try simplifying things one piece at a time. Note that $\displaystyle\sum_{j = 1}^{n}a_{ij}x_j = (Ax)_i$ (the $i$-th component of $Ax$). Thus, $\displaystyle\sum_{i = 1}^{m}a_{ik}\left(\sum_{j = 1}^{n}a_{ij}x_j - b_i\right) = \sum_{i = 1}^{m}a_{ik}((Ax)_i-b_i) = \sum_{i = 1}^{m}a_{ik}(Ax-b)_i = (A^T(Ax-b))_k$ Therefore, $\displaystyle\sum_{i = ...


2

This is a special case of the tensor product of vectors known as outer product. It has some interesting properties, for example that the trace of the matrix is the square of the (Euclidean) norm of the vector. And as you point out it is always symmetric.


2

The result will be $A^TA$ (because $IA = A$, of course, even for rectangular matrices). And $A^TA$ will be a diagonal matrix with the diagonal entries being the square of the diagonal entries of $A$ (and some zeroes, if $m < n$). Proof Let $A = [a_{ij}]_{m \times n}$. Then $A^TIA = A^TA = B = [b_{ij}]_{n \times n}$ (say). $$\begin{align} b_{ij} & = ...


2

Given $n+1$ points ${\bf a}_k$ $(0\leq k\leq n)$ in ${\mathbb R}^n$ one considers their convex hull $S$ which is a (maybe degenerate) $n$-dimensional simplex. This simplex is spanned by the $n$ vectors ${\bf a}_k-{\bf a}_0$ $(1\leq k\leq n)$, and its volume is equal to the volume of the parallelotope $P$ spanned by these vectors, divided by $n!$ (see below). ...


2

Hints rather than answers: (a) how 'big' is the subspace spanned by the last three rows? Can three vectors be linearly independent in that subspace? (b) Are you familiar with the pigeonhole principle? This is (at some level) a very simple application of that, and related to the answer to (a): each of the 120 terms in the determinant corresponds to one of ...


2

Hint: If a matrix $M$ is written in square blocks, like $$M = \begin{pmatrix} A & B \\ \mathbf{0} & D \end{pmatrix}$$ then: $$\det M = \det A ~\det D$$


2

Some of the properties of the matrix you are talking about are In my area of research (signal processing), this is referred to as the gram matrix. I believe, this is the standard name as well. It is a positive (semi) definite (PD) matrix. If you know about PD matrices, proving this is straight forward. The trick is this relation $x^Tx=||x||_2^2\geq 0$. ...


2

Hint for a): What does it mean for $z$ to be an eigenvector? It means that $Az=\lambda z$. When would you say that two complex quantities are equal? You equate some parts. What, with what? Hint for b) Assume they are linearly dependent and $y=kx$ for some $k$. Then use (a) and the fact that $b \neq 0$.


2

The triangular matrices are not normal in general. Only the diagonal matrices are. So take a conjugate of $$A=\begin{bmatrix} 0 & 1 & 1\\0 & 0 &1\\0 & 0 &0\end{bmatrix}$$ For example, consider $PAP^{-1}$ with $$P=\begin{bmatrix} 1 & 0 & 0\\0 & 1 &1\\1 & 0 &1\end{bmatrix}$$ which happens to be ...


2

To construct examples of this, it's useful to take the $2\times 2$ case and extend it to larger sizes First observe that there are no diagonalizable matrices with this property. Thus, such matrices, if they exist are similar to either $\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\end{bmatrix}$ or $\begin{bmatrix}0 & 1 ...


2

(a) The matrix $A$ is nilpotent so it isn't diagonalizable unless $A=0$ (b) The polynomial $x^2-1$ with simple roots $\pm1$ annihilates $A$ then (b) is true (c) isn't true since $x^2-x$ has simple real roots so it's diagonalizable over $\Bbb R$. Take for example a projection. (d) isn't true for example any nilpotent matrix has the same characteristic ...


2

Notice that if you apply the cofactor expansion along the first column we have a lot of zeroes so we only need to worry about $3$ multiplied by the determinant of a smaller matrix. Notice that there is another candidate column to which to apply the cofactor expansion in the smaller matrix (the column with the most zeroes). Continue this process. Notice that ...


2

Multiplying a diagonal matrix by itself, $A^2$, will result in each of the diagonal entries being squared. For example, your matrix $A$ has diagonal entries 1, 2, and 4. $A^2$ has entries 1, 4, and 16. The trace of this matrix is 21.


2

No. In the case of square matrices, all the other row operations either leave the determinant unchanged or negates it. So if the original matrix has determinant $\notin\{-1,0,1\}$, then its reduced row echelon form is the identity matrix (with determinant $1$) but cannot be reached using the other row operations.


2

Evaluating $P$ at the given vector tells you two things: first, $a-b=-3$, and second that the given vector is an eigenvector with eigenvalue $3$. The trace of the matrix is $1+b$, but it is also the sum of the eigenvalues, so $1+b=6$, and $b=5$. Since $a=b-3$, $a=2$.



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