let$ A = \begin{bmatrix}2& −2& −4\\−1 &3& 4\\1 &−2 &-3\end{bmatrix}$
What is the minimal polynomial of A?
i find the eigenvalue of A that is $\lambda = 1, 0$
minimal polynomial of A = $(\lambda -1 )\lambda$
EDits : minimal polynomial of A = $(\lambda -1 )^2\lambda$
Is it true ????
see my edits
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The minimal polynomial has the property that when you substitute the matrix for the variable, you get zero. Is it the case that $(A - I)A = 0$?
This criterion is not strong enough to show you are done. You also have to show that no divisor of a polynomial giving you zero is a polynomial that gives you zero. (Much as the minimal polynomial divides the characteristic polynomial, both of which give zero when the matrix is substituted for the variable.)
No, the fact that the eigenvalues if $A$ are $0$ and $1$ is not enough to prove that its minimal polynomial is $\lambda(\lambda-1)$. For instance, the minimal polynomial of\begin{pmatrix}1&1&0\\0&1&0\\0&0&0\end{pmatrix}is $\lambda(\lambda-1)^2$.
However, since$$\bigl((4,0,1),(2,1,0),(1,-1,1)\bigr)$$is an ordered basis of $\mathbb{R}^3$, since its first two vectors are eigenvectors of $A$ with eigenvalue $1$ and since the third one is an eigenvector of $A$ with eigenvalue $0$, $A$ is similar to$$\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$and therefore, yes, its minimal polynomial is $\lambda(\lambda-1)$.