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9

In a quite pedestrian way, this is just saying that $\det$ is a quadratic form with the trace as its polar form. Namely, for any matrix $A$, let $A^\star = \mathrm{Tr} A - A$ be its conjugate (as in “the conjugate root of the characteristic polynomial”). Then your formula is equivalent to $$ \det (A+B) - \det(A) - \det(B) = \mathrm{Tr}(A^\star \cdot B).$$ ...


8

If $A$ is $m\times n$ real matrix and $\operatorname{rank}A = k\le\min(m,n)$, then $AA^T$ is $m\times m$ matrix of rank $k$. Thus $AA^T$ is invertible if and only if $\operatorname{rank}A = m$ (in particular: if $m > n$, then $AA^T$ must be singular.) For complex matrices it does not hold in general, as baharampuri suggests. However for complex matrix ...


6

This is the case $n=2$ of a 1980 theorem of Amitsur. It is described (in more abstract notation) here and this question gives the reference to Amitsur's paper.


6

Laplace expansion: Let $A$ be an $n \times n$ matrix. Denote $a_{ij}$ be the entry of row $i$ and column $j$. Denote $A_{ij}$ be the $(n-1) \times (n-1)$ matrix by deleting row $i$ and column $j$. Then $\det A$ is given by: $$\det A = \sum^n_{j=1}(-1)^{i+j}a_{ij}\det A_{ij}$$ Picture: Intuitively, the signs are alternating in this pattern: ...


5

Instead of memorizing it, we can express the determinant of a $n \times n$ matrix by: $$\det(A) = \sum_{\sigma \in S_n} {\rm sgn}(\sigma) \prod_{i=1}^n a_{i,\sigma(i)}.$$ The above is often used as the definition of the determinant. For the case $n=2$, for example, $$\sigma_1 = {\rm id},\, {\rm sgn}(\sigma_1)=1, \quad \sigma_2 = (1\, 2), \, {\rm ...


5

A nice intuitive visualization is: The determinant is the (oriented) $n$-dimensional volume of the epiped ("distorted cube") spanned by the column vectors. Especially, the deteminant is zero iff the epiped is "squashed" to lower dimensions, i.e., iff the matrix is not invertible.


5

I think you may have miscalculated your determinant. I get $$-\lambda^3 + 6\lambda^2 - \frac{15}{2} \lambda + \frac{5}{2}$$ for which you can check 1 is a root (usually a guess worth trying). I suppose you can proceed from there. Also, a handy way to check consistency is to see if the trace $\text{Tr}(A)$ equals the $\lambda^2$ coefficient, since they both ...


5

The matrix is the special case of so-called Vandermonde Matrix (See Induction for Vandermonde Matrix). det($A$)$=(a-b)(a-c)(b-c)$. So if $a\neq b,a\neq c, b\neq c, \space$ det$(A)\neq 0$.


4

The calculation of the determinants does involve basic properties of the determinant function: Multilinearity: Let $\alpha \in \mathbb{R}$ and $A$ an $n \times n$ matrix. Then $$\det( \alpha A) = \alpha^n\det(A).$$ Multiplication of determinants: For two $n \times n$ matrices $A,B$ $$\det(A B ) = \det(A) \cdot \det(B).$$ Let now $A$ be an ...


4

Hint For any $n \times n$ matrices $A, B$, we have $$\det(AB) = \det(A) \det(B).$$ Any scalar matrix $\lambda I$ is diagonal, so its determinant is the product of its diagonal entries: $$\det(\lambda I) = \lambda^n.$$


4

Hint $A$ has positive eigenvalues because it is diagonally dominant (why?) and the diagonal entries are positive. This suffices to show that $\det A = \prod_{i=1}^n \lambda_i > 0$ where $\lambda_i$ are the eigenvalues of $A$.


3

This matrix has determinant $1$, see here. The matrix looks as follows $$ \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \cr 1 & 2 & 2 & \cdots & 2 \cr 1 & 2 & 3 &\cdots & 3 \cr \vdots & \vdots & \vdots & \ddots & \vdots \cr 1 & 2 & 3 & \cdots & n \end{pmatrix} $$


3

Note that $\det B = 1$. Note additionally that $B^{-1} = A(A^T)^{-1}$, so that $$ \operatorname{trace}(B^{-1}) = \operatorname{trace}(A(A^T)^{-1}) = \operatorname{trace}((A^T)^{-1}A) = \operatorname{trace}(B^T) = \operatorname{trace}(B) $$ Now, your polynomial can be written as $$ \det(I + B) = \det(B) + \det(B)\operatorname{trace}(B^{-1}) + ...


3

I am not sure that this is what you want, but it is the cleanest solution I can find. It is easy to check that, if all but one of the $v_j$ are fixed and both the function is viewed as a function of a single vector, it will be a linear function. Additionally, if two of the $v_j$s are the same, say $v_k=v_{\ell}$ the left hand side will be zero. To see ...


3

Here's an answer that completely avoids determinants. Determinants are heinously overrated. Let $\vec v_1,\vec v_2,\dotsc,\vec v_n$ be the columns of a matrix $A$. That is, $$ A= \begin{bmatrix} \vec v_1 & \vec v_2 & \dotsb & \vec v_n \end{bmatrix} $$ Now, suppose the columns of $A$ are not linearly independent. Then there exist scalars ...


3

As Matt Samuel noted, this is because we use multilinear forms to measure area. Why multilinear forms? Because this takes into account not just area but orientation. Let's take a little jaunt through what an area function should be and why this involves multilinear algebra, and we'll finish with the general formula for a determinant. Properties of an area ...


3

Call $a= \det A$. Then $aI$ is a diagonal matrix, so its determinant is the product of its diagonal entries. So $$\det( (\det A)I) = \det(aI) = a^n = (\det A)^n$$


3

Hint : The minimal polynomial of the matrix $A$ must be a divisor of the polynomial $$x^3-x^2-3x+2=(x-2)(x^2+x-1)$$


3

No. The operation you carried out is not true. Though you got the number of real roots as 2, which is correct, but the real roots are incorrect. The roots of the equation are 2,-2. You can either split open one row or a column but not all in the same step. For a speedy way to solve it: Take $x^2=t$ and then expand. It will take you less than a ...


2

$\begin{vmatrix}x-1&x+3&2\\-3&x+5&-6\\x+3&2&x+k\end{vmatrix}$ If $(x-2)$ is a factor of this then the determinant $\Delta(x)$ should vanish at $x=2$ that means $\Delta(2)=0$ $\begin{vmatrix}1&5&2\\-3&7&-6\\5&2&2+k\end{vmatrix}=0$ $C_3\to C_3-2C_1$ ...


2

You cannot use the ``rule'' $\det (A+B) = \det A + \det B$, as it's just wrong :) Try adding the second row to the first and subtracting the third row from it. Such row operations do not change the determinant.


2

Suppose $v$ is an eigenvector of $A$, with eigenvalue $\lambda$. Then $\lambda$ must satisfy the same cubic equation that $A$ satisfies: $$(A^3-A^2-3A+2I)v=0v=0\\ (A^3-A^2-3A+2I)v=AAAv-AAv-3Av+2v\\ =AA\lambda v-A\lambda v-3\lambda v+2v\\ =\lambda A(Av)-\lambda(Av)-3\lambda v+2v\\ =\lambda A(\lambda v)-\lambda^2 v-3\lambda v+2v\\ ...


2

$$B=\begin{bmatrix} n-1 & -1 & \cdots & -1 \\ -1& n-1 & \cdots & -1\\ \vdots&\vdots &\ddots &\vdots \\ -1& -1 &\cdots & n-1 \end{bmatrix}=nI_{n-1}-\begin{bmatrix}1\\1\\\vdots\\1\end{bmatrix}\cdot\begin{bmatrix}1&1&\cdots&1\end{bmatrix}=nI_{n-1}-A$$ Now, $\text{rk}A=1$ and ...


2

Let \begin{eqnarray}v_i=(a_1^i, a_2^i, a_3^i, b_i)\end{eqnarray} That the three determinants are 0 implies that the row vectors are linearly dependent. In particular, $\text{dim Span}\{v_1, v_2, v_3, v_4\}\leq 3$, $\text{dim Span}\{v_2, v_3, v_4, v_5\}\leq 3$, $\text{dim Span}\{v_3, v_4, v_5, v_6\}\leq 3$. So \begin{eqnarray}\text{dim Span}\{v_1, v_2, v_3, ...


2

First note that the determinant is cyclic. Hence, it is of the form $f(a,b,c)$, where $f$ is a polynomial of degree $5$. Further, we have $f(a,a,c) = f(a,b,b) = f(c,b,c) = 0$, which means $(a-b)$, $(b-c)$ and $(c-a)$ are factors, i.e., the determinant is $g(a,b,c)(a-b)(b-c)(c-a)$, where $g(a,b,c)$ is a cyclic polynomial of degree $2$. Any cyclic polynomial ...


2

If I've well understood, the matrix is as follows: $$A-\lambda I=\begin{bmatrix} 1-\lambda&1&0&0&\dots &0\\ 0&-\lambda&1&0&\dots&0\\0&0&-\lambda&1&\dots&0\\\vdots&\vdots&\vdots&\vdots&&\vdots\\ 0&0&0&0&\dots&1\\ 1&0&0&0&\dots&-\lambda ...


2

The wronskian is a function, not a number, so you don't can't say it's lower or higher than $0(x)$. You may get either $ g(x) $ or $-g(x)$ depending on row placement but it matters little. You only care about whether or not said $ g(x) $ is 0 for all x.


2

The importance of the Wronskian is the fact that its non-zero can tell you about linear independence (if it is zero, you need some additional conditions to say something about linear dependence). The "sign" doesn't matter -- multiplying any of the functions you're testing by $-1$ would also change the sign of the determinant but lead you to the same logical ...


2

This is A180128 in the OEIS. You can do better for $n=4:$ $$ \pmatrix{53&11&23&13\\ 17&47&29&3\\ 7&5&43&37\\ 19&31&2&41}=4868296>4673460. $$ The fifth term is $$ \pmatrix{89&41&23&2&53\\ 31&97&29&47&11\\ 59&13&79&61&7\\ 37&19&5&83&67\\ ...


2

How would this (or an alternate intuitive demonstration) be extended to a three dimensional case, where the formula is the following and calculates volume? Yes. Just as the determinant of the $2\times 2$ matrix is the area of the parallelogram formed with the two column vectors as sides, so too is the determinant of a $3\times 3$ matrix the area of a ...



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