Skip to main content

New answers tagged

1 vote
Accepted

Biot-Savart Law Application Simplification Issues

$\newcommand{\bvec}[1]{\mathbf{#1}}$ $\newcommand{\uv}[1]{\bvec{\hat{e}}_{#1}}$ Work in the traditional Heaviside–Gibbs vector calculus with cylindrical coordinates $(\rho,\phi,z)$. Let $\bvec{J}$ ...
K B Dave's user avatar
  • 8,166
0 votes

$A^{p^{n!}}-A$ is nilpotent in $M_n(\mathbb{F}_p)$

This answer is based on Daniel Schepler's comments. Let $A=PJP^{-1}$ where $J$ is the Jordan Normal form of $A$ over $\mathbb{F}_{p^{n!}}$. Note that actually entries of $P$ are also in $\mathbb{F}_{p^...
Laurence PW's user avatar
0 votes

Linear Algebra and Multivariable Calculus book/lectures after AP Calculus BC

Let me share some of my experiences with you. I also did AP Calculus BC (got a 5 btw). Because of that I was (also) able to start with Vector Calculus (50A,50B) at Berkeley. Well, as you can ...
soft tostada burrito's user avatar
6 votes

Can there be a perfect linear operator for square matrices?

If you don't want to invoke ring theory: Let $n\geq 2$. For any such $f$, we have that if $A^2=A$, then $f(A)=0$ or $f(A)=1$. If $f(I_n)=0$, then $f$ is constant $0$, so we may assume that $f(I_n)=1$....
Arturo Magidin's user avatar
6 votes

Can there be a perfect linear operator for square matrices?

All such maps are either the constant zero map or an algebra homomorphism from the real algebra $M_n(\mathbb R)$ to the real algebra $\mathbb R$. Over a field, matrix algebras are simple. Thus, this ...
NoOneIsHere's user avatar
0 votes

Find the inellipse of a triangle with its major axis pointing in a given direction and with a given ratio of major to minor

Following the idea presented by @David K in his comment above, I will first transform the given triangle vertices by scaling by $r = \dfrac{b}{a}$ along the direction $V = (\cos \theta , \sin \theta) ...
i don't know what i am doing's user avatar
13 votes
Accepted

Can there be a perfect linear operator for square matrices?

Edit: for completion, I added the $n=1$ and odd $n$ cases, using the helpfull comments of Will Jagy. First, observe that $f$ is a linear map between vectors spaces $ \mathbb{R}^{n \times n}$ and $\...
smitke6's user avatar
  • 484
5 votes
Accepted

True or False: Inner product on $\mathbb{R}^2$ satisfying a specific norm.

The actual parallelogram law is an identity, not an inequality: $$\|v+y\|^2+\|v-y\|^2=2\|v\|^2+ 2\|y\|^2$$ and that is not satisfied in your example, hence the norm is not associated with an inner ...
Lieven's user avatar
  • 648
2 votes

Comprehensive Linear Algebra Text

caffeinemachine's comment recommends Linear Algebra by Kenneth Hoffman and Ray Kunze (first published by Prentice-Hall in 1961), but it seems that only Pearson Education India still reprints this book....
Tsundoku's user avatar
  • 207
2 votes

Exercise books in linear algebra

There are several options in addition to those already suggested in other answers: Luis Barreira, Claudia Valls: Exercises In Linear Algebra. World Scientific, 2016. Each chapter contains a ...
Tsundoku's user avatar
  • 207
1 vote

If $A_{3\times 3}$ is a matrix whose row sums are 1, and $\det(A)\neq 0$.Also, $B=I+A^{-1}+A^{-2}+A^{-3}$, then row sums of matrix B is?

Let $e$ be the vector of ones. You are given that $Ae=e$ and since $A$ is invertible, $A^{-1} e = e$, repeating shows that $A^k e = e$. Hence $Be = 4e$ and hence the row sums are all $4$.
copper.hat's user avatar
  • 173k
3 votes

$\operatorname{rank}(A)=r$, then $I,A,\cdots, A^{r+1}$ is linearly dependent

You can do a bit better: without using $I$, those positive powers of $A$ are already linearly dependent. Let $V$ be the image (column space) of $A$, then by definition of the rank one has $\dim(V)=r$, ...
Marc van Leeuwen's user avatar
1 vote
Accepted

Range of values of θ that satisy these values.

The system of equations can be rewritten as $$\begin{pmatrix} 1& \sin(\theta)\\ \sin(\theta) & 4 \end{pmatrix}\begin{pmatrix} x\\ y \end{pmatrix}=\begin{pmatrix} 1\\ 2 \end{pmatrix}$$ ...
Julio Puerta's user avatar
  • 7,552
0 votes

Non-zero generalized eigenvalues of permuted accretive matrix

After spending 30 hours on this problem before posting, i found the solution within 20 minutes... thats just classic. Anyway, heres the answer: Assume $Nv = 0$ and $(\lambda-JA)^{2n}v = 0$. Claim: $$\...
Emzatin's user avatar
2 votes
Accepted

Minimizing the trace of matrix using matrix calculus

$\def\tr{\operatorname{tr}}$ $\def\qty#1{\left(#1\right)}$ To understand this think of the following simple problem. You are given a matrix, $$ \mathbf{X}= \begin{bmatrix} x & y \\ z & t \end{...
Ted Black's user avatar
  • 750
0 votes

eigenvalue of $X \mapsto A X B$

As Gunnar mentioned in comments, this is not true in general, without assuming $\mathbb{F}$ is algebraically closed. When $\mathbb{F}$ is algebraically closed (say, $\mathbb{F} = \mathbb{C}$) though, ...
David Gao's user avatar
  • 6,380
1 vote

Discriminant formula - do the coefficients include their signs?

Just rewrite everything in general form $$ ax^2+bx+c=0 $$ You can use round brackets if helpful $$(a)x^2+(b)x+(c)=0$$ Eg1 $$ 2x^2 - 2x = 0 $$ Rewrite $$0 = 2x^2 - 2x$$ $$= (2)x^2 - 2x$$ $$= (2)x^2 + -...
BCLC's user avatar
  • 13.5k
0 votes

Give an example of $T \in \mathcal{L}(\mathbb{R^2})$ such that $T^4=-I$.

Hint for a geometric method The map $-I$ geometrically represents a reflection through the origin, or just as well as a rotation by $\pi$. So for any positive integer $n$, the (say, anticlockwise) ...
Travis Willse's user avatar
0 votes

Coordinates, components, and change of basis

Having read and I think mostly followed your thoughts here, I believe that what's happened is this: your confusion begins quite a bit earlier than you may actually realize. You have accepted as true ...
john's user avatar
  • 2,396
0 votes

Coordinates, components, and change of basis

Since you rightfully mentioned the chain rule I demonstrate (again) in detail how straightforward it is to derive everything about an arbitrary coordinate transformations from that. The coordinate ...
Kurt G.'s user avatar
  • 14.8k
1 vote
Accepted

linear functions in a bounded region

You don't need to consider the $x$ or $y$ axis in any of this, but your thought about lines of constant value is good. Consider, in particular, the line $f(x, y) = f(A)$, i.e. the line of constant ...
ConMan's user avatar
  • 24.8k
0 votes

Give an example of $T \in \mathcal{L}(\mathbb{R^2})$ such that $T^4=-I$.

Building off of what J.W. Tanner stated in their answer, one could cite that $T$ need only be a member of the set $\mathrm{SO}(2)$. Specifically, $T$ may be a 2-dimensional rotation matrix which ...
Arjun Vyavaharkar's user avatar
2 votes

Give an example of $T \in \mathcal{L}(\mathbb{R^2})$ such that $T^4=-I$.

Hint: Think of rotations. $-I$ represents a $180^\circ$ rotation. Can you think of rotations that become a $180^\circ$ rotation when repeated four times?
J. W. Tanner's user avatar
1 vote
Accepted

Orthogonal projection is bounded

First you need the fact that $P^*=P$. Indeed, $$ \langle P^*(u+w),u+w\rangle=\langle u+w,P(u+w)\rangle=\langle u+w,u\rangle=\langle u,u\rangle=\langle u,u+w\rangle=\langle P(u+w),u+w\rangle. $$ It ...
Martin Argerami's user avatar
0 votes
Accepted

Bound on minimum perturbation of eigenvalues, based on condition number

Figured I'd finally give this question a fully-fleshed answer in order to close it. This essentially expounds upon the proof in the Wikipedia link. As noted in the comments, this result is ...
PrincessEev's user avatar
  • 44.3k
1 vote
Accepted

how does a row or column of 0s affect if a matrix is defective?

The matrix $A$ is clearly defective. This is because, it is in its Jordan form with only two Jordan blocks, as opposed to $5$ different blocks for diagnolizability. It is not row or column of zeros ...
vidyarthi's user avatar
  • 7,067
2 votes

Minimizing the trace of matrix using matrix calculus

$ \def\a{\alpha} \def\o{{\tt1}} \def\BR#1{\left[#1\right]} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\frob#1{\left\| #1 \right\|_F} \def\q{\quad} \...
greg's user avatar
  • 36.4k
1 vote

($X^{T}AX=0,\;\;\forall X$) iff $A$ is a skew symmetric matrix

Seven years later... The “only if” part can be proved in a relatively clean way via the polarisation identity. Given that $x^TAx=0$ for all vectors $x$, we have, for all $u$ and $v$, \begin{align*} 0&...
user1551's user avatar
  • 140k
0 votes

($X^{T}AX=0,\;\;\forall X$) iff $A$ is a skew symmetric matrix

For every A and x we know $x^TAx=(x^TAx)^T$ $$A=-A^T$$ $$x^TAx=-x^TA^Tx=-(Ax)^Tx=-(x^TAx)^T$$ therofore $x^TAx=0$
Reza's user avatar
  • 1
1 vote

Is there a surjective group homomorphism $\operatorname{GL}_{n}(k) \to \operatorname{GL}_{m}(k)$ where $n > m$?

Here is an alternative approach, which may require some refining. Assume $n>m$ and we have a group homomorphism $\phi:\mathrm{GL}_n(k)\to\mathrm{GL}_m(k)$ with non-abelian image. There is a non-...
Andrew Hubery's user avatar
4 votes
Accepted

Discriminant formula - do the coefficients include their signs?

Yes, pay attention to the sign. The quadratic formula gives you an explicit expression for the solutions to the quadratic polynomial equation $$ ax^2 + bx + c = 0, \tag{1} $$ where $a$, $b$, and $c$ ...
Sammy Black's user avatar
  • 25.7k
2 votes

Discriminant formula - do the coefficients include their signs?

It is $-2$ . Does the coefficient count with the positive/negative sign? Yes the coefficient count with the sign. In your case the discriminant would be: $D = (-2)^2 -4(2)(0)$ $D = 4 - 0$ $D = 4$ ...
madhurkant's user avatar
2 votes

Discriminant formula - do the coefficients include their signs?

General If, the general form is $$ax^2+bx+c=0$$ Then, for any value of $b$, the coefficient of the equation must be with the sign. Specific Answer The values of the coefficients are : $a=+2$, $b=-2$
aditya's user avatar
  • 79
7 votes

Discriminant formula - do the coefficients include their signs?

Considering the general form $$ ax^2+bx+c=0 $$ $$\implies b=-2$$ Since $$x^2-2x=x^2+(-2)x$$ The discriminant would be the same even if $b=2$, but $x$ will be affected, this is because if we look at ...
Masd's user avatar
  • 884
6 votes
Accepted

Can the expression $\frac{x^{T}Ax}{x^{T}Bx}$ be re-written in the form $\frac{x^{T}B^{-1}Ax}{x^{T}x}$?

First, optimizing $\frac{x^T A x}{x^T B x}$ is not the same as optimizing $\frac{x^T B^{-1}A x}{x^T x}$. For a specific example, consider $$A = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix} , \...
lisyarus's user avatar
  • 15.7k
1 vote
Accepted

Line through $A$ parallel to $\alpha$ and incident on $r$.

Indeed, there is no solution. I follow your proof until equation $$x+y+z=3\tag{1}$$ Afterwards, I would re-write it a little differently as follows. (1) means that any line passing through $A$ and ...
Jean Marie's user avatar
  • 82.9k
6 votes

Does anyone know a non-trivial surjective multiplicative homomorphism from the $4\times 4$ matrices to the $2 \times 2 $ matrices?

If there is a surjective multiplicative homomorphism $\varphi$ from $M_r(k)$ to $M_s(k)$ where $r>s>1$, then $\varphi(I_r)$ is a multiplicative identity on the range of $\varphi$, hence $\...
Just a user's user avatar
1 vote
Accepted

Intersection of two vector parametric equations of lines

First of all, consider the definition of being perpendicular. Given your vector $\mathbf{v}$, a vector $\mathbf{u} = (a, b)^T$ is perpendicular to $\mathbf{v}$ if the inner product between the two is ...
rcescon's user avatar
  • 264
9 votes

Does anyone know a non-trivial surjective multiplicative homomorphism from the $4\times 4$ matrices to the $2 \times 2 $ matrices?

Claim: There exists no surjective multiplicative morphism from $M_r(k) \to M_s(k)$ for $r > s > 1$. Proof: Assume the contrary, let $\varphi: M_r(k) \to M_s(k)$ be such a morphism. For all $A \...
Haran's user avatar
  • 10k
0 votes

Non-negativity constraints in linear-programming formulation of $L_1$ norm minimization

It is not necessarily true that $x \ge 0$. Consider the one-dimensional example of minimizing $|x+1|$. That is, $n=1$, $A=(1)$, and $b=-1$. The unique optimal solution is $x=-1$, and the ...
RobPratt's user avatar
  • 46k
3 votes

Why is (I+A) invertible and what is its inverse?

Under some conditions on a real number $x,$ we have $$ \frac{1}{1+x} = 1 - x + x^2 - x^3 + x^4 - ... $$ Note that any power of a square matrix $A$ commutes with any other power. And Under some ...
Will Jagy's user avatar
  • 140k
0 votes

What is so special about eigenvector that it behaves like a pivot of linear transformation?

I suppose one way to look at it is that it's the vector that has the right combination of $\hat{\imath}$ and $\hat{\jmath}$ such that the stretching and skewing caused by $A$ on each of those basis ...
ConMan's user avatar
  • 24.8k
0 votes
Accepted

Find subspace $T$ of $\mathbb{R}^{3}$ such that $\mathbb{R}^{3} =V \oplus T$

Actually, $T=\Bbb R^3$, and therefore it is false that $V\cap T=\{0\}$. Note that $(1,2,1)=\frac12(2,8,10)-(0,2,4)$. Since, on the other hand, $\{(2,8,10),(0,2,4)\}$ is a linearly independent set, $V=\...
José Carlos Santos's user avatar
0 votes

Find subspace $T$ of $\mathbb{R}^{3}$ such that $\mathbb{R}^{3} =V \oplus T$

First determine the dimension of $V.$ The 3 vectors that generate it, are not linearly independent; if they were, then the dimension would be 3 and $T$ could only be the null space. As it is, try and ...
Lieven's user avatar
  • 648
1 vote

Continuous basis $(e_1(t),e_2(t))$

This can be proved with elementary linear algebra. Let us write $A_t$ for $A(t)$. Let $r$ be the constant rank of all $A_t$, where $0<r<n$. (In your case $r=n-2$.) For any pair of matrices $M$ ...
user1551's user avatar
  • 140k
2 votes
Accepted

Inverse of the identity minus a block anti-diagonal matrix.

The matrix $I-C$ is a block matrix with 2 invertible matrices on the diagonal. You can use the equation that you are linking to by substituting an $I$ of the appropriate size for $A$ and $D$, $-A$ for ...
Lieven's user avatar
  • 648
0 votes

Is every vector space an injective module?

For any vector space $M$, $$\mathrm{Ext}^1_F(M,Q)=0$$ since $M$ is free, hence projective. Thus, $Q$ is injective.
Juan L.'s user avatar
  • 635
0 votes

How to identify surfaces of revolution

This would be an answer for Computer Graphics/Computer Scientist folks: Check the following recent paper which provides algorithmic tools for identifying patches of surface of revolution on ...
2 votes
Accepted

automorphisms on direct sum of matrices

The centre $Z$ of $M_n(R)\oplus M_m(R)$ is $\def\espan{\operatorname{span}}Z=\espan\{I_n\oplus 0, 0\oplus I_m\}$. Because $f$ is an automorphism, $f(Z)=Z$ (proof at the end). As $f(0)=0$ and $f(I)=I$, ...
Martin Argerami's user avatar
0 votes
Accepted

eigenvalue bounds

Use the Eigenvalue interlacing theorem. See Theorem 1 of the attached link. https://people.orie.cornell.edu/dpw/orie6334/Fall2016/lecture4.pdf
Debarghya's user avatar
  • 135

Top 50 recent answers are included