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Example of real matrices $2\times 2$ and $3\times 3$ that are not similars to a diagonal matrix.

I find that $A =\begin{bmatrix} 2 & 1 \\ 0 & 1 \end{bmatrix} $ then i suppose that its similar to some diagonal matrix and found $D = \begin{bmatrix} 2 & 0 \\ 0 & 2\end{bmatrix} $ = $2I$ which is not the case.

But I can't find a real matrix $3 \times 3$ that is not similar to a diagonal matrix. Some help for this please.

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  • $\begingroup$ $A$ has two distinct eigenvalues, so it's similar to a diagonal matrix; your $D$ is diagonal, so it's similar to a diagonal: itself. I suggest you to look at $\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$ and try to generalize to a $3\times 3$ matrix. $\endgroup$ – egreg Jun 17 '14 at 17:39
  • $\begingroup$ but they have the same characteristic equation. Hence A and D have the same eigenvalues. Since the eigenvalues of D of the numbers on the diagonal, and the only eigenvalue of A is 2, then we must have as D, In this case, we must have $A = P^{-1}DP = 2 I$, which is not the case. Therefore, A is not similar to a diagonal matrix. $\endgroup$ – Knight Jun 17 '14 at 17:41
  • $\begingroup$ Of course not; the characteristic polynomial of $A$ is $(2-x)(1-x)$, the characteristic polynomial of $D$ is $(2-x)^2$. The matrix $A$ is similar to $\begin{bmatrix}2&0\\0&1\end{bmatrix}$. $\endgroup$ – egreg Jun 17 '14 at 17:43
  • $\begingroup$ o yea your right, i will try you example $\endgroup$ – Knight Jun 17 '14 at 17:50
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Since the eigenvalues of $\begin{bmatrix}a&b\\0&c\end{bmatrix}$ are $a$ and $c$, and having $n$ distinct eigenvalues for an $n$-dimensional transformation implies the transformation is diagonalizable, you'll need to look at cases where $a=c$.

Then, say, we could try $\begin{bmatrix}1&b\\0&1\end{bmatrix}$, which just has $1$ as an eigenvalue. If $b=0$ this is obviously diagonalizable, so say $b\neq 0$.

The equation $\begin{bmatrix}1&b\\0&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x+by\\y\end{bmatrix}=\begin{bmatrix}x\\y\end{bmatrix}$ would imply $by=0$, hence $y=0$. Then the only eigenvectors possible are $\begin{bmatrix}x\\0\end{bmatrix}$, which only furnishes a $1$ dimensional eigenspace for the value $1$.

Keeping in mind that diagonalizable transformations must have a basis consisting of eigenvectors, we can see that is not possible here, so the matrix isn't diagonalizable.

Try to do something similar for $3\times 3$ matrices. Using just $1$ and $0$ and keeping the matrices as simple as possible will allow you to argue in a simliar way.

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