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26

Hint: look at the sum of the last two rows.


7

Hint. Start with $$\det{\begin{bmatrix}A & B \\-B & A\end{bmatrix}}=\det{\begin{bmatrix}A-iB& B+iA \\-B & A\end{bmatrix}}.$$


5

You will always get $\det(BA)=0$. The reason for this is very simple : $BA$ is a $3\times 3$ matrix with rank at most $2$; thus it is not invertible and thus it has $0$ determinant. One possible definition of the rank is that it is the dimension of the subspace generated by the columns or the lines of the matrices. The lines of $BA$ are obtained by taking ...


5

We are assuming that $A$ is real and $n\times m$. Suppose that $\det(A^TA)=0$. Then there exists nonzero $v\in \mathbb R^n$ such that $A^TAv=0$. Then $$ 0=A^TAv=v^TA^TAv=(Av)^TAv. $$ For any real vector $w$, if $w^Tw=0$, then $w=0$. Thus, $Av=0$. But this is exactly, if $C_1,\ldots,C_m$ are the columns of $A$, $$ 0=v_1C_1+v_2C_2+\cdots+v_mC_m. $$ So the ...


5

Since the determinant of a matrix $M$ is the product of the eigenvalues of $M$, this yields an empty product for $det([\ ])$, which is by definition 1. The reason the empty product is defined to 1 (this is one reason of many) is because it makes the definition of exponentiation in terms of multiplication work. Since $$a^n = a * a * \ldots * n \ times$$ And ...


5

Let us call $M$ the initial matrix. The answer is: if $n \geq 3$, $\det(M)=0$ . The proof is as follows. Let $$M_1=\begin{bmatrix}e^{ix_1}\\e^{ix_2}\\\vdots\\e^{ix_n}\end{bmatrix}\begin{bmatrix}e^{-iy_1}&e^{-iy_2}&\cdots&e^{-iy_n}\end{bmatrix}$$ It is a rank 1 matrix. Therefore one can write $$M=\dfrac{1}{2}(M_1+\bar{M_1})$$ Thus, as the ...


4

Since $A(\operatorname{com}A)^T = \det (A) I_n$, $$\det (A) \det (\operatorname{com}A) = (\det(A))^n$$ Now it remains to check cases, whether $A$ is invertible or not. If $\operatorname{rank} A =n$, $A$ is invertible and $\det (\operatorname{com}A) = (\det(A))^{n-1}$ If $\operatorname{rank} A = n-1$, $A(\operatorname{com}A)^T=0$. As a result, ...


4

Here's a quicker argument. You know $A$ and $B$ are invertible, so $$ABA = A$$ implies $AB = I$ by multiplying $A^{-1}$ on the right. However, your argument is correct. Edit: After thinking about this more, I think the point of the original exercise is to prove that every square matrix with a left inverse also has a right inverse and vice versa. I suppose ...


4

We can prove a much more general (and useful) result. Theorem. The (real) matrices $A$ and $A^TA$ have the same null space, hence the same rank. (I'll use $N(B)$ to denote the null space of $B$: $N(B)=\{v:Bv=0\}$.) Proof. It is obvious that $N(A)\subseteq N(A^TA)$. Suppose $v\in N(A^TA)$; then $A^TAv=0$ implies $v^TA^TAv=0$, so $(Av)^T(Av)=0$, hence ...


4

The matrix can be expressed as the sum of two rank $1$ matrices, each of which is the outer product of vectors $$M=\mathbf{11}^T+\mathbf{xy}^T$$ where $\mathbf{1}=[1,1,1,\cdots,1]^T\in\mathbb{R}^{n\times 1}$, $\mathbf{x}=[x_1,x_2,\cdots,x_n]^T\in\mathbb{R}^{n\times 1}$ and $\mathbf{y}=[y_1,y_2,\cdots,y_n]^T\in\mathbb{R}^{n\times 1}$. Using the subadditive ...


4

I would discourage you from using the word "adjoint" in this context. This is an accepted usage of the word, but there is another concept in linear algebra which is always referred to by the word "adjoint". The two can be easily confused. An unambiguous word that can be used in this context is "adjugate", and I would encourage you to use this word. Anyway, ...


4

$$\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right| =\left|\begin{matrix}1&a&a^2\\0&b-a&b^2-a^2\\0&c-a&c^2-a^2\end{matrix}\right| =\left|\begin{matrix}b-a&b^2-a^2\\c-a&c^2-a^2\end{matrix}\right|=(b-a)(c-a)\left|\begin{matrix}1&b+a\\1&c+a\end{matrix}\right|=(b-a)(c-a)(c-b).$$ ...


3

The only linear map $\{0\}\to\{0\}$ is the identity map and the identity map has determinant $1$, because this is a required property of the determinant. From a slightly different point of view, since the basis of $\{0\}$ is empty, the representative matrix of the unique endomorphism is the empty matrix (which is invertible and its inverse is, of course, ...


3

Let $\left( \begin{matrix} x \\ y \end{matrix} \right) \in \mathbb{R}^{2n}$ be an eigenvector with eigenvalue $\lambda$. Then $\left( \begin{matrix} -y \\x \end{matrix} \right)$ is also an eigenvector with eigenvalue $\lambda$, so all eigenvalues have an even number of linear independent eigenvectors.


3

Consider two cases. $\def\rk{\operatorname{rank}}\rk([A~B])<n$. Then $\det(A)=\det(B)=0$ and the result is obvious. $\rk([A~B])=n$. Now the fact that adding the rows of $[C~D]$ to the matrix does not increase the rank means the every such row is a linear combination of the rows of $[A~B]$. This means that there is some $n\times n$ matrix $X$ such that ...


3

well, a diagonal matrix with nonzero entries along the main diagonal is always invertible, so to answer the original question there is no rule of thumb. Furthermore, a matrix that is all nonzero may be non-invertible. You would need to look at the rank of the matrix (or many, many other things, like the determinant)


2

I will only give the results and the strategy here, since the calculation is a little bit complicated. Using the laplacian method you can calculate the determinant along the first column. Using the property that changing two columns or rows of a matrix the determinant only changes its sign there are actually three parts: ...


2

Suppose that $b,c\ge 0$. Since $u=(a+d)/2$, we have $$ad-bc=\left(\frac{a+d}{2}\right)^2+v^2,$$ Multiplying the both sides by $4$ gives $$4ad-4bc=a^2+2ad+d^2+4v^2$$ i.e. $$-4bc=(a-d)^2+4v^2$$ The LHS is non-positive since $b,c\ge 0$, and the RHS is positive since $v\not=0$. This is a contradiction. Hence, either $b$ or $c$ has to be negative.


2

Given an irreducible degree $2$ monic polynomial over the reals it is possible to find a $2\times 2$ matrix having the given polynomial as its characteristic polynomial. Take $x^2-2x+2$ as the polynomial. A matrix having it as its characteristic polynomial is the “companion matrix” $$ \begin{bmatrix} 0 & -2 \\ 1 & 2 \end{bmatrix} $$ The matrix has ...


2

Let $ T = BA $, then $ T : \mathbb{R}^3 \to \mathbb{R}^3 $ is a linear transformation with nontrivial kernel, as $ A : \mathbb{R}^3 \to \mathbb{R}^2 $ cannot be an injection. Thus, it is not invertible, and $ \det(T) = 0 $.


2

Notice that if we set $x=-b$, then our determinant $\Delta(-b)$ is upper-triangular, and we have $\Delta(-b) = P(b)$. Likewise, with $x=-a$, we have $\Delta(-a) = P(a)$. Use these two facts, along with the result $$\Delta(x)=Ax+B$$ which you previously obtained to simplify the expression $$\frac{aP(b) - bP(a)}{a-b},$$ and you find precisely that it's ...


2

Take the transpose of $$ \left(\begin{array}{ccccccc} 1 & -1 & 0 & 0 & \ldots & 0 & 0 \\ 2 & -1 & 0 & 0 & \ldots & 0 & 0\\ 3 & 0 & 1 & 0 & \ldots & 0 & 0 \\ 4 & 0 & 0 & 1 & \ldots & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots ...


2

HINT.- Part (a) are simple exercises of matrix multiplication you have to learn. Part (b), because of the multiplication of a matrix by a scalar, can be possible if and only if there is compatibility of the four following equations: $$2a+b=1\\1\cdot a+0\cdot b=-2\\3a+0\cdot b=-6\\a+b=3$$ You can verify whitout difficulty that it is in fact compatible.


2

If $x=y=1,\ z=0.1$ the determinant is $-0.3957...$ And for $x=y=1,z=0.5$ the determinant is $+0.1781...$ So since it is continuous in $z$ for fixed $x,y$ there is a value of $z$ between $0.1$ and $0.5$ for which $(x,y,z)=(1,1,z)$ makes the determinant zero. Maybe the ordering in one of the rows should be different, but this can only be checked from the ...


2

I am interpreting the matrix $A$ to be a sequence of matrices, whose limit is taken entrywise, and $B$ likewise. Consider $A_j=\left(\begin{smallmatrix}j&0\\0&1\end{smallmatrix}\right)$. We have $det(A_j)=j$, which approaches infinity as $j\to\infty$. However, $A_j^{-1}=\left(\begin{smallmatrix}\frac{1}{j}&0\\0&1\end{smallmatrix}\right)$, ...


2

A non-constant entire function that has constant modulus on the unit circle must be a constant multiple of the power function. It follows that $\det(zI+AB^{-1})=z^n$ and $AB^{-1}$ is nilpotent. As $B$ is invertible and it commutes with $A$, $A$ must also be nilpotent. When $A,B$ do not commute, the assertion does not necessarily hold. It is easy to ...


1

Hint. By continuity, you may assume without loss of generality that $A$ is invertible. Then, use Schur complement to show that the determinant is equal to $\det(A)^2\ \left|\det(I+iBA^{-1})\right|^2$.


1

Denote the determinant by $f(x)$. Then the $x$ coefficient is given by $f'(0)$. Now the derivative of a determinant equals the sum of determinants of the matrices obtained by taking the derivative of each row separately. Hence $$ \begin{align} f'(0) &= \frac{\mathrm{d}}{\mathrm{d}x} \left| \begin{matrix} (1+x)^{a_1b_1} & (1+x)^{a_1b_2} & ...


1

If you take that equation seriously for $k=1$, you need to evaluate $\det A_0$, the determinant of an empty matrix. That's the sum over an empty product for all permutations of $0$ elements, of which there is $1$, so $\det A_0=1$. Thus the case $\det A_k\lt0$ is excluded. A more direct way might be to consider that equation to hold for $k\gt1$ and regard the ...


1

Since the columns of $A$ are linearly dependent then $A$ is injective. In fact if $Ae_1,\ldots,Ae_n$ are the columns of $A$ and if $x=\sum_{i=1}^nx_ie_i\in \ker A$ then $0=Ax=\sum_{i=1}^n x_i Ae_i\implies x_i=0\implies x=0$. Now if $\det(A^TA)=0$ then $0$ is an eigenvalue of $A^TA$ and let $x$ an associated eigenvector. Then $$0=\langle A^TAx,x\rangle=\Vert ...



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