Consider the matrix: $$ A=\left(\begin{matrix} a & a & a \\ a & a & a \\ a & a & a \end{matrix}\right) $$ How to prove whether $A$ is diagonalisable or not. And how to diagonalize it.
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1Possible duplicate of Matrix with all 1's diagonalizable or not? – Dietrich Burde May 13 '19 at 16:25
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What is the standard procedure to find the Eigenelements ? – May 13 '19 at 16:25
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3$A$ has eigenvalues $3a$, and $0$ of algebraic multiplicity $2$. Since the geometric multiplicity of $0$ eigenvalue is also $2$, the matrix is diagonalizable (assuming over $\mathbb C$ for instance). – Dave May 13 '19 at 16:28
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@Dave thanks, is there a way to diagonalize it using elementary operations on rows and colones? – David Lingard May 13 '19 at 16:40
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To diagonalize, first notice that $$A\begin{bmatrix}1\\1\\1\end{bmatrix}=\begin{bmatrix}3a\\3a\\3a\end{bmatrix}$$ so $$u:=\begin{bmatrix}1\\1\\1\end{bmatrix}$$ is an eigenvector of the $3a$ eigenvalue. To find eigenvectors $v,w$ of the $0$ eigenvalue, simply reduce $A$ to its RREF and determine a basis for the kernel. Then we can diagonalize $A$ as $A=P^{-1}DP$ with $$D:=\begin{bmatrix}3a&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ and $P:=\begin{bmatrix}u&v&w\end{bmatrix}$.
Dave
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Thanks, do elementary operations also work to transform it to a diagonal matrix? What I did is this $ A=\left(\begin{matrix} a & a & a \ 0 & 0 & 0 \ 0 & 0 & 0 \end{matrix}\right) $ but I stopped at this point. – David Lingard May 13 '19 at 17:52
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I'm not sure what you're trying to use elementary operations to do (i.e. what do you mean by "transform it to a diagonal matrix"?). – Dave May 13 '19 at 17:55
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I mean using Gaussian elimination, that is using elementary row operations to modify the matrix until we find a diagonal matrix (the same procedure we use to find the inverse of a matrix) In the end we find a diagonal matrix. But is this diagonal matrix $D$ similar to the original matrix $A$ ($A=P^{-1} D P$)? – David Lingard May 14 '19 at 00:28