# Questions tagged [jordan-normal-form]

The Jordan normal form of a matrix is a similar block matrix having diagonal blocks when the matrix is diagonalizable and diagonal + nilpotent blocks more generally.

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### Prove that $V = \ker(\phi^n) \oplus \text{image}(\phi^n)$

Let $V$ be a $n$-dimensional complex vector space and $\phi:V\to V$ a linear mapping. Prove that $$V = \ker(\phi^n) \oplus \text{image}(\phi^n)$$ Here is my attempt: Since $\phi^n$ is also a linear ...
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### Finding Jordan form

Find Jordan form of the following matrix: $$\left(\begin{matrix}4&-5&2 \\ 5&-7&3\\ 6&-9&4 \end{matrix}\right)$$ So I got stuck pretty much trying to find the eigenvalues. ...
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### Non-integral powers of a matrix

Question Given a square complex matrix $A$, what ways are there to define and compute $A^p$ for non-integral scalar exponents $p\in\mathbb R$, and for what matrices do they work? My thoughts ...
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### Finding the Jordan canonical form of this upper triangular $3\times3$ matrix

I am supposed to find the Jordan canonical form of a couple of matrices, but I was absent for a few lectures. \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 3 \end{...
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### Size of Jordan block

Imagine that I'm writing the Jordan form of a matrix and I know that the eigenvalue needs to appear 4 times in the diagonal (algebraic multiplicity is 4) and we need 2 Jordan blocks (geometric ...
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### $n$-th root of $3 \times 3$ invertible matrix

Yo, I couldn't solve this exercise after thinking for a while. For every $A \in GL_{3} (\mathbb{C})$ and $n$, there's a $B \in Mat_{3, 3}(\mathbb{C})$ such that $B^n = A$ The previous exercise was ...
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### How to find the 'real' jordan canonical form of a matrix

Given that the the Jordan normal form of a matrix is, $J=\begin{bmatrix}2&1&0&0\\0&2&0&0\\0&0&1-i&0\\0&0&0&1+i\end{bmatrix}$ How do you find the 'real'...
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### Prove that $A$ is similar to $A^n$ based on A's Jordan form

Let $A = \begin{bmatrix}1&-3&0&3\\-2&-6&0&13\\0&-3&1&3\\-1&-4&0&8\end{bmatrix}$, Prove that $A$ is similar to $A^n$ for each $n>0$. I found that ...
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### On the rank inequality $\operatorname{rank}(A)+\operatorname{rank}(A^3)\geq2\operatorname{rank}(A^2)$

So, I have to show that for all square matrices $A$, $$\newcommand{\rank}{\operatorname{rank}}\rank(A)+\rank(A^3)\geq2\rank(A^2).$$ My attempt at it so far: So, if I try to use this theory: Let $A$ ...