Diagonalisation Test

Topics: Algebra - Diagonalisation of a Linear Transformation - Linear Transformation - Vector Space

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Let $T:V\to V$ with $\dim (V)=n$.

$T$ is diagonalisable if:

1. $f(t)=(-1{)}^n(t-{\lambda }_1)\dots (t-{\lambda }_n)$
2. The multiplicity of $\lambda$ is equal to $n-\text{ran}(T-\lambda I)$ for each eigenvalue

Example

Let’s see if $A=\left(\begin{array}{ccc}3& 1& 0\\ 0& 3& 0\\ 0& 0& 4\end{array}\right)$. is diagonalisable.

First condition §

We calculate the characteristic polynomial:

$$\det (A-\lambda I)=\det \left(\begin{array}{ccc}3-\lambda & 1& 0\\ 0& 3-\lambda & 0\\ 0& 0& 4-\lambda \end{array}\right)$$

$$=(3-\lambda {)}^2(4-\lambda )$$

$$=(-1)(\lambda -3{)}^3(\lambda -4)$$

With that, we get that ${\lambda }_1=4$ (with a multiplicity of 1) and that ${\lambda }_2=3$ (with a multiplicity of 2).

Since we get that $f(t)=(-1)(\lambda -3)(\lambda -3)(\lambda -4)$, then the first condition is satisfied.

Second condition §

As for the second condition, we have to check for every eigenvalue.

First eigenvalue §

First, we have that ${\lambda }_1$ has a multiplicity of 1. Let’s calculate $\text{ran}(A-{\lambda }_{1}I)$:

$$\text{ran}(A-{\lambda }_{1}I)=\text{ran}\left(\begin{array}{ccc}-1& 1& 0\\ 0& -1& 0\\ 0& 0& 0\end{array}\right)=2$$

To find the range of a matrix, we counted the amount of columns that are linearly independent.

With that, we do get that $\dim ({\mathbb{R}}^3)-\text{ran}(A-{\lambda }_{1}I)=3-2=1=$ the multiplicity of $l_1$. The second condition is satisfied with the first eigenvalue.

Second eigenvalue §

We do the same we did for the first eigenvalue:

$\text{ran}(A-{\lambda }_{2}I)=\text{ran}\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right)$

This time, we get that $\dim ({\mathbb{R}}^3)-\text{ran}(A-{\lambda }_{2}I)=3-2=1\ne$ the multiplicity of $l_1$. The second condition is not satisfied with the second eigenvalue.

Since the second condition is not satisfied in this case, then it follows that $A$ isn’t diagonalisable.