What is the proof for this and the intuitive explanation for why the reduced row echelon form have the same null space as the original matrix?

A group $G$ is generated by $\begin{pmatrix}1&n\\0&1\end{pmatrix}$ and $\begin{pmatrix}1&0\\n&1\end{pmatrix}$, then we know $G\cong \mathbb{F}_2$ which is a free group generated by two elements. Now I consider the representation: $G\to GL(2,\mathbb{R})$, it is necessary that the image of $\begin{pmatrix}1&n\\0&1\end{pmatrix}$ is a triangular matrix under conjugation? Thanks in advance.

[Ciarlet 1.2-2] Let $O$ be an orthogonal matrix. Show that there exists an orthogonal matrix $Q$ such that $$Q^{-1}OQ\ =\ \left(\begin{array}{rrrrrrrrrrr} 1 & & & & & & & & & & \\ & \ddots & & & & & & & & & \\ & & 1 & & & & & & & […]

I want to prove (or disprove) the following statement: If $A$ is a square matrix with non-negative elements that has spectral radius less then $1$, then any matrix obtained from $A$ by arbitrarily changing the sign of the elements has the same property. This problem appeared recently when studying the convergence of some matrices and […]

This question is motivated by this question, which gave me quite a headache today. Context: I posted originally what I thought was a quick proof using the derivative of the given function. It was intended to be a straightforward answer using a well-known strategy of proving something is constant and equal to some other thing […]

In order to find the inverse matrix $A^{-1}$, one can apply Gaussian-Jordan elimination to the augmented matrix $$(A \mid I)$$ to obtain $$(I \mid C),$$ where $C$ is indeed $A^{-1}$. However, I fail to see why this actually works, and reading this answer didn’t really clear things up for me.

Given a singular matrix $A$, find the eigenvalues of the adjugate matrix of $A$. The same question with $A$ being invertible is trivial since $A\operatorname{adj}A=(\operatorname{adj}A)A=(\det A) I$. If $\operatorname{rank}A\leq n-2$, it is well-known that $\operatorname{adj}A=0$ and $0$ is the only eigenvalue. It remains to deal with the case $\operatorname{rank}A= n-1$. It is easy to check […]

I was kinda crushed to discover that two different matrices with different properties can actually share the same characteristic polynomial ($-\lambda^3-3\lambda^2+4$): $A=\begin{pmatrix} 1 & 2& 2\\ -3 &-5 &-3 \\ 3& 3 & 1 \end{pmatrix} , B=\begin{pmatrix} 2 & 4& 3\\ -4 &-6 &-3 \\ 3& 3 & 1 \end{pmatrix}$ $A$ has an eigenline and […]

Consider the matrix \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{matrix} what effect does $({x_1},{y_1})$,$({x_2},{y_2})$,$({x_3},{y_3})$ being collinear have on the rank of the above matrix ?

The definition of the adjugate matrix is easy to understand, but I have never seen it used for anything. What is the intuitive meaning of this matrix? Are there examples of applications which may shed light on its conceptual meaning? I would be especially interested to hear examples of usage in representation theory or other […]

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