Tagged Questions

For any topic related to matrices. This includes: systems of linear equations, eigenvalues and eigenvectors (diagonalization, triangularization), determinant, trace, characteristic polynomial, adjugate, transpose, Jordan normal form, matrix algorithms (e.g. LU, Gauss elimination, SVD, QR), ...

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If $AB = I$ then $BA = I$

If $A$ and $B$ are square matrices such that $AB = I$ where $I$ is identity matrix. Show that $BA = I$. I do not understand anything more than the following. Elementary row operations. Linear ...
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Similar matrices and field extensions

Given a field $F$ and a subfield $K$ of $F$. Let $A$, $B$ be $n\times n$ matrices such that all the entries of $A$ and $B$ are in $K$. Is it true that if $A$ is similar to $B$ in $F^{n\times n}$ then ...
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Given a $4\times 4$ symmetric matrix, is there an efficient way to find its eigenvalues and diagonalize it?

I have a $4\times 4$ matrix $$A=\left(\begin{array}{cccc}8 & 11 & 4 & 3\\11 & 12 & 4 & 7\\4 & 4 & 7 & 12\\3 & 7 & 12 & 17\end{array}\right).$$ I want to ...
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Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!)

First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an ...
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Why do the $n \times n$ non-singular matrices form an “open” set?

Why is the set of $n\times n$ real, non-singular matrices an  open subset of the set of all $n\times n$ real matrices? I don't quite understand what "open" means in this context. Thank you.
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Proof of this result related to Fibonacci numbers: $\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}$?

$$\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}$$ Somebody has any idea how to go about proving this result? I know a proof by ...
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Intuition behind Matrix Multiplication

If I multiply two numbers, say $3$ and $5$, I know it means add $3$ to itself $5$ times or add $5$ to itself $3$ times. But If I multiply two matrices, what does it mean ? I mean I can't think it ...
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System of linear equations having a real solution has also a rational solution.

I saw this question Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $b ∈ \mathbb{Q}^m$. Suppose that the system of linear equations $Ax = b$ has a solution in $\mathbb{R}^n$. Does it necessarily have ...
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Calculate Rotation Matrix to align Vector A to Vector B in 3d?

I have one triangle in 3d space that I am tracking in a simulation. Between time steps I have the the previous normal of the triangle and the current normal of the triangle along with both the current ...
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Looking for an intuitive explanation why the row rank is equal to the column rank for a matrix

I am looking for an intuitive explanation as to why/how row rank of a matrix = column rank. I've read the proof at http://en.wikipedia.org/wiki/Rank_of_a_linear_transformation and I understand the ...
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When is matrix multiplication commutative?

I know that matrix multiplication in general is not commutative. So, in general: $A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A$ But for some matrices, this equations holds, e.g. A = ...
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Why, historically, do we multiply matrices as we do?

Multiplication of matrices — taking the dot product of the $i$th row of the first matrix and the $j$th column of the second to yield the $ij$th entry of the product — is not a very ...
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Is the rank of a matrix the same of its transpose? If yes, how can I prove it?

I am auditing a Linear Algebra class, and today we were taught about the rank of a matrix. The definition was given from the row point of view: "The rank of a matrix A is the number of non-zero ...
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Why is the complex number $z=a+bi$ equivalent to the matrix form $\left(\begin{smallmatrix}a &-b\\b&a\end{smallmatrix}\right)$ [duplicate]

Possible Duplicate: Relation of this antisymmetric matrix $r = \left(\begin{smallmatrix}0 &1\\-1 & 0\end{smallmatrix}\right)$ to $i$ On Wikipedia, it says that: Matrix ...
Under what conditions does a matrix $A$ have a square root? I saw somewhere that this is true for Hermitian positive definite matrices(whose definition I just looked up). Moreover, is it possible ...