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1

Only square matrices have determinants. I'm not sure whether your matrix was square to begin with, but it's not square both before and after adding that extra row. If $Ax=0$ has a unique solution, $B$ is the matrix obtained from $A$ by adding another row, and that row is a linear combination of the rows of $A$ then yes, $Bx=0$ has a unique solution. In fact ...


1

I don't have Hoffman & Kunze at hand, but the definition of row echelon form (which is embedded in your definition of reduced row echelon form) from wikipediae states: The leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the leading coefficient of the row ...


1

If you take an initial point $p_i$, apply a transform $A$, then apply transform $B$, then apply transform $C$ to get a final point $p_f$, then we have $$p_f = C(B(A(p_i)))$$ $$p_f = CBAp_i$$ So the total transformation is $t = CBA$. This is one of the reasons why matrix multiplication is so nice--it lets us compose many linear transforms into a single ...


0

While a span is always a subspace and not linearly independent, a basis may not be a subspace and is always linearly independent.


1

Each 4×4 transformation matrix is of the form $$ T_i = \begin{vmatrix} R_i & \vec{t}_i \\ 0 & 1 \end{vmatrix} $$ such that when applied to a 4×1 point $(\vec{p},1)$ the result is a rotation and a translation $(\vec{t}_i+R_i \vec{p},1)$ Now combine three transformations for $$\begin{vmatrix} R_1 & \vec{t}_1 \\ 0 & 1 \end{vmatrix} ...


1

It's probably because it's not a geometrically meaningful operation; a linear transformation whose matrix in one basis is all ones, has another matrix in another basis. Whenever I've seen the notation $A+b$ in mathematics, it has meant $A+bI$ (where $A$ is a quadratic matrix and $I$ is the identity matrix of the same size). Some people write ...


1

Yes, that relationship is true. $A^2$, $A$, and $I$ can all be simultaneously diagonalized by an orthogonal matrix $E$. If $\Lambda$ are the eigenvalues of $A$ then we know that $$A^2 = E\Lambda^2 E^{-1} \,\,\,\, A = E\Lambda E^{-1} \,\,\,\, I = E I E^{-1}$$ The question of whether the ordering you specified is true is by definition the question of whether ...


0

Hint: Let's call your three points $a$, $b$, and $c$. Let's call the center of the circle $p$. First, let's assume that $a$, $b$, and $c$ are NOT the same line, 'co-linear'. That needs to be handled separately because $p$ cannot be not defined in that case. We know that $p$ is equidistant from $a$, $b$, and $c$ - because that's the defining property of a ...


2

The first question only requires the definition of an eigenvalue. For the second one, since $g\in \mathbb{C}[X], g(X)=c\prod\limits_{i=1}^n(X-\lambda_i)$ so $g(A)=c\prod\limits_{i=1}^n(A-\lambda_i I)$. And since the product of invertible matrices is invertible, you can conclude using the first point.


0

By Brauers Theorem ( Matrix Analysis by Horn & Johnson ) subtracting by $\lambda {\bf I}$ will reduce all eigenvalues by $\lambda$ and if none equals $\lambda$ means each of them will become non-zero.


1

One way is to use eigen decomposition. If A is diagonalizable matrix then $ A = Q^{-1} \Lambda Q $ where $\Lambda$ is diagonal matrix with eigevalues on the diagonal. Then $ A^k = Q^{-1} \Lambda^k Q $ and $g(A) = Q^{-1} g(\Lambda) Q $. $g(\Lambda)$ is diagonal matrix with $g(\lambda_i)$ elements. From this place your proposition is easy to prove. ...


0

Hint: $$g(x)\in \mathbb{C}[x]\implies \sigma\left(g(A)\right)=\{g(\lambda):\lambda\in \sigma(A)\}$$ where $\sigma(\cdot)$ denotes the spectrum, i.e. collection of eigenvalues of a matrix.


0

Note that the element $M_{11}$(also called the first minor of the matrix) is positive, which is stated in the question. In order for this matrix to be negative definite, the determinant (or equivalently the second minor $M_{22}$ needs to be positive. If it is zero then the matrix is negative semidefinite. If it's, on the other hand, negative then it is ...


0

Counterexamples: - with $A=I_2$ the identity matrix and $\Sigma^{-\frac{1}{2}}=\operatorname{diag}\left(\frac{1}{2},2\right)$ we get $A^{\top}A=I_2$ and $B^{\top}B=\operatorname{diag}\left(\frac{1}{4},4\right)$. Thus, both matrices have different eigenvalues. - with $A=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)$ also all eigenvectors ...


0

Of course $\lambda\not=\lambda_1$. By the power method, you can obtain the greatest eigenvalue $\lambda_2=tr(B^TB)-\lambda_1 $ of $B^TB$ and an associated eigenvector $v_2$. Let $P=[v_1,v_2]\in M_2$; then you know $B^TB=Pdiag(\lambda_1,\lambda_2)P^{-1}$; finally you must remove the scaling: $A^TA=\Sigma^{1/2}Pdiag(\lambda_1,\lambda_2)P^{-1}\Sigma^{1/2}$.


18

Note that applying function to a matrix is meant in sense of series, that is if $$ \phi(z) = c_0 + c_1 z + c_2 z^2 + \dots $$ then $$ \phi(\mathbf A) = c_0 \mathbf I + c_1 \mathbf A + c_2 \mathbf A^2 + \dots. $$ Observe that $$ \phi(z) = \frac{e^z - 1}{z} = 1 + \frac{z}{2} + \frac{z^2}{6} + \dots = \sum_{k = 1}^\infty \frac{z^{k-1}}{k!}. $$ Also, to your ...


8

Note that since $$ \exp(A)=I+A+\frac1{2!}A^2+\frac1{3!}A^3+\cdots $$ then $A$ and $\exp(A)$ commute. Thus also $\exp(A)-I$ and $A^{-1}$ commute.


1

If you look at the series expansion of this function at $z=0$, you find there are only positive powers of $z$. In any case, naively $\frac{1}{z}=z^{-1}$ just translates to the inverse, which exists for many matrices. For all these translations, note, however, that $xy=yx$ need not hold for matrices. But if a function only involves one matrix variable ...


3

Try writing out the series for $exp$ and simplify the equation before you evaluate! You wouldn't have to worry about dividing!


0

As Travis pointed out, by polarizing we have that: $\langle Qx, Qy \rangle$=$\langle x,y\rangle \quad \forall x,y$ Hence: $\langle Q^TQx, y \rangle$=$\langle x,y\rangle \quad \forall x,y$ $\implies \langle Q^TQx-x, y\rangle=0 \quad \forall x,y$ Fix an arbitrary $x$. Put $y:=Q^TQx-x$. Therefore, $\langle Q^TQx-x, Q^TQx-x\rangle=0 \implies Q^TQx-x=0 ...


0

Let me establish the question again here: Let $A \neq 0, I$ be a $n \times n$ matrix and $P$ be a $n \times m$ matrix. Suppose $P = AP$. Can we conclude $P = 0$? The answer is generally no. Consider $A = \begin{pmatrix} 1&0\\0&0 \end{pmatrix}$ and $P = \begin{pmatrix}1\\0\end{pmatrix} = e_1$. Then $P = e_1$ is an eigenvector of $A$ with respect ...


1

Suppose $Qx \cdot Qx = x \cdot x, \forall x \in \Bbb R^n; \tag{1}$ then $x^TQ^TQx = Qx \cdot Qx = x \cdot x$ $= x^T Ix, \forall x \in \Bbb R^n, \tag{2}$ whence, $x^T(Q^TQ - I)x = 0, \forall x \in \Bbb R^n; \tag{3}$ we note that $Q^TQ - I$ is symmetric: $(Q^TQ - I)^T =(Q^TQ)^T - I^T = Q^TQ - I. \tag{4}$ The desired result now follows with the aid of ...


3

The matrices $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ can be obtained from one another by a row operation, yet one is diagonal and the other not diagonalizable. Note that row operations "destroy" the identification between domain and target, which is essential to the notion of diagonalizabilty.


0

Consider the orthonormal basis $\{e_i\}_{i=1}^n$ and consider a matrix $Q=(Q_{ab})$ that satisfies $(2)$. Then, for $1\leq i,j\leq n$, we have $$ Qe_i\cdot Qe_j = e_i^TQ^TQe_j= \sum_{1\leq a,b\leq n} Q_{ib}^TQ_{aj} = e_i\cdot e_j = \delta_{ij}. $$ On the other hand, note that $$ \left(Q^TQ\right)_{ij} = \left(QQ^T\right)_{ij} = \sum_{1\leq a,b\leq n} ...


1

Hint Polarizing the equation in (2) gives that $$Qx \cdot Qy = x \cdot y$$ for all $x, y \in \Bbb R^n$, and using the same manipulation as in the proof of the other implication gives that this is equivalent to $$y^T Q^T Q x = y^T x.$$ Now, evaluate both sides at $x = e_a$, $y = e_b$, where $(e_1, \ldots, e_n)$ is the standard basis.


0

Given your current problem statement, it is not clear that $W^{-1}$ exists. For example, if $U$ is $n \times n$, then it is possible that $W$ is $n \times k$ and $X$ is $k \times n$, where necessarily $k \geq n$ in order for both sides of the matrix equation $U = WX$ to have full rank. Since transposing the problem converts you from rows into columns, you ...


1

It has been shown that the determinant is something like: $$V_{k+1}(x) = f(x) = C(x-a_2)(x-a_3)\ldots(x-a_k)(x-a_{k+1})$$ Note that if you would expand this it would be of the form: $$f(x) = Cx^k + B_{k-1}x^{k-1}+\ldots$$ Now consider the expansion theorem for the first row: $$V_{k+1}(x) = \begin{vmatrix} a_2^{k-1} & \cdots & a_2^2 & a_2 ...


1

Take the column $k$. If we denote $x^k=z$ then it looks like $$ \left[\matrix{z-1\\ z^2-1\\z^3-1\\ \vdots\\ z^n-1}\right]=(z-1)\left[\matrix{1\\ z+1\\z^2+z+1\\ \vdots\\ z^{n-1}+\ldots+z+1}\right]. $$ Now use the top row of ones to eliminate all other ones and $z$ to eliminate all other $z$ and so on $$ (z-1)\left[\matrix{1\\ z+1\\z^2+z+1\\ \vdots\\ ...


0

It's talking about how you work out determinants you can "expand" them into a sum of 2x2 determinants It's what I did here ( Vandermonde determinant by induction ) 2 years ago.


0

Burnside's theorem. Let $K$ be an algebraically closed field. The algebra generated by $A,B\in M_n(K)$ is whole $M_n(K)$ iff $A,B$ have no common invariant proper subspace. Then, your assertion is false, except if $n=2$. EDIT. (Answer to Mariano). If $n=2+p\geq 3$, then let $K^n=E_2\oplus F_p=E_2\oplus G_p$ where $F_p\cap G_p=\{0\}$. Let $A$ be $E_2$ and ...


1

Call your matrix $A$. If $X=(a,b,c,\ldots,f)^T$ is a column vector then the polynomial $$ p(y) = ay+by^2+cy^3+\cdots+fy^n-(a+c+d+\cdots+f) $$ applied to $x, x^2, x^3, \ldots, x^n$ will produce the elements of $AX$. If $AX=0$, then accordingly $x, x^2, x^3, \ldots, x^n$ are all roots of $p$. We can also see directly that $1$ is a root of $p$. If the powers ...


0

There are several mistakes in the body of question. $\nabla(f)(x)=2Ax-b$. Assume that $A$ is a square matrix. Then $x^*$ is a critical point of $f$ iff $(A+A^T)x^*=b$; as Thoth wrote, putting $A:=A+A^T$, we may assume that $A$ is symmetric, $f(x)=1/2x^TAx-b^Tx$ and the required equation becomes (*) $Ax^*=b$. 2 $A$ must be invertible, otherwise there is ...


1

if you choose any nilpotent matrix M so you have existence of $k\in\mathbb{N}$ such that $M^k=0$ so in $M^n=0$ for all $n\geq k$ and : $$ \|M^n\|^{1/n}=0 $$ So $\rho(M)=\lim_n \|M^n\|^{1/n}=0$. we can conclude that the spectral radius of any nilpotent matrix is equal to $0$. and so the spectral radius can't be a norme.


1

Choose matrices x and y such that the fourth property (triangle inequality) is not satisfied. There are plenty of matrices which will show that $$\rho(A+B) > \rho(A)+\rho(B)$$


1

Deleted my first post since it looks so lengthy. Straightforward I.E. here. The end matrix will look like the 2 by 3 matrix there. So let's compute the upper lefthand number. That number, 0, is in the first row and first column. So the row we are working with is the first row: $$\begin{pmatrix}-1 & 3 \end{pmatrix}$$ The column is the first column: ...


1

First, note that the general form of $\bf M$ is $$\pmatrix{a & \mp b \\ b & \pm a},$$ that is, e.g., if one takes $-b$ for the $(1, 2)$ entry, one must take $+a$ for the $(2, 2)$ entry. (One can show this for example by writing a general matrix $A$ and determining the algebraic conditions on the entries determined by the defining condition $A^T A = ...


0

If the first matrix had a third column, then you wouldn't be able to multiply. "Row times column" requires that the numbermof columns in the first matrix agrees with the number of rows in the second one.


0

$AB$ is defined whenever $A$ is $m \times k$ and $B$ is $k \times n$ for some integers $k,m,n$. The resulting matrix $C\equiv(c_{ij})$ is $m\times n$. In particular, $c_{ij}=\sum_{k} a_{ik}b_{kj}$. These matrices are $2 \times 2$ and $2 \times 3$. Can you use the above to come to a conclusion?


0

Begin by multiplying everything in matrix A by 2. Then add them respectively to B. 2A = \begin{bmatrix} 4 & 6\\ 4 & 8\\ \end{bmatrix} B = \begin{bmatrix} 0 & 4\\ -1 & 6\\ \end{bmatrix} 2A+B= \begin{bmatrix} 4 & 10\\ 3 & 14\\ \end{bmatrix}


0

Do you mean: $$\mathbf{M} = \begin{bmatrix} a & \pm b \\ b & \mp a \end{bmatrix}, |\operatorname{det}(\mathbf{M}) = |a^2 + b^2| = 1$$ ? To complete the proof, note that $|a^2 + b^2|= 1$ imply that $a \in [-1,1] $, so there exists such a $\theta \in [0,2\pi)$, that $\cos(\theta)=a$. Next, by the same equality, you have: ...


1

You are performing operations between matrices and the result is a matrix. Your two operations are: 1) multiplication by a scalar, defined as $$ k \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ \end {bmatrix}= \begin{bmatrix} ka_{11}&ka_{12}\\ ka_{21}&ka_{22}\\ \end {bmatrix} $$ 2) sum of matrices, defined as $$ \begin{bmatrix} ...


0

The answer below is correct, but I just want to be sure you know that multiplication of a matrix by a scalar (in this case 2) gives you another matrix. Also, the addition of two matrices gives you another matrix. What you have proposed would just give you a number when you need the result to be a matrix.


1

Multiply $2$ to matrix $A$, which is just multiply every number in $A$ by $2$. Then add the numbers in $A$ to $B$ respectively, that is, i. e. upper left corner number + upper left corner number. Here's the 'upper left corner number' done (you can work out the rest yourself, better practice anyways) $2\cdot 2 = 4$. Add to matrix $B$: $$4 + 0 = 4.$$ Your ...


1

Note that $$\begin{align}\chi_{A^2}(t^2)&=\det(A-t^2I)\\&=\det((A-tI)(A+tI))\\&=\det(A-tI)\det(A+tI)\\&=\chi_A(t)\chi_A(-t)\end{align}$$ so that (at least in characteristic $0$) $$ \chi_{A^2}(\lambda)=\chi_A(\sqrt\lambda)\chi_A(-\sqrt\lambda)$$ If we know the factorization $$\chi_A(\lambda)=\prod_k(\lambda-c_k)$$ the we obtain ...


0

We have $\begin{bmatrix}-5& -9\\ 9 & 3 \end{bmatrix} + \begin{bmatrix}8& 5\\ -4 & -1\end{bmatrix} -\begin{bmatrix} 4& -7\\ -9 & -6 \end{bmatrix} =\begin{bmatrix} (-5+8-4)& (-9+5+7)\\ (9-4+9) & (3-1+6) \end{bmatrix} =\begin{bmatrix} -1& 3\\ 14 & 8 \end{bmatrix}.$


1

If a matrix $A$ is invertible, then $A\vec x=\vec b$ has a unique solution $\vec x=A^{-1}\vec b$. By the rank theorem we know that $rank(A)+nullity(A)=n$. If $rank(A) < n$, then $nullity(A) \geq 1$. But this means that $A\vec x = \vec 0$ has infinitely many solutions. This is not possible if $A$ is invertible, because we expect a unique solution.


2

The rank of a matrix is the number of linearly independent column vectors of this matrix, i.e. the dimension of the subspace $\,\operatorname{Im}A$. Furthermore, $A$ can be seen as the matrix associated, in a certain basis, to a linear map $f\colon K^n\to K^n$ ($K$ is the base field). For an $n\times n$-matrix, this means that ...


1

You did actually everything alright, but as already pointed out in the comment $$ v=\begin{pmatrix}-8\\7\\12\end{pmatrix} $$ is simply not an eigenvector to the eigenvalue $\lambda=-2$. You can choose for example $$ \hat{v}=\begin{pmatrix}-1\\2\\3\end{pmatrix} \in ker(A+2I) $$ and then everything works flawless, then you can further take ...


4

The rank of a matrix is equivalent to the number of nonzero rows after Gaussian elimination. If the rank is not maximal, then there is at least one nonzero row after elimination. Do the determinant about this row, and it will be $0$, i.e., the matrix is not invertible.


2

Let's look at the $3 \times 3$ case in general. If $A = [A_1|A_2|A_3]$ then note for $c = [c_1,c_2,c_3]^T$: $$ Ac = [A_1|A_2|A_3]\left[ \begin{array}{c} c_1 \\ c_2 \\ c_3 \end{array} \right] =c_1A_1+c_2A_2+c_3A_3.$$ Thus, the equation $Ac=0$ is equivalent to the condition $c_1A_1+c_2A_2+c_3A_3=0$. Now, if $\{A_1,A_2,A_3 \}$ is a linearly independent set then ...



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