# How to get the eigenvalue of this cases

I'm in a Linear Algebra class, where we are currently covering eigenvalues and eigenvectors. My question is if my answer is correct.

Consider the matrix $$\begin{equation*} A := \begin{bmatrix} 0 & 0 & a \\ b & c & 10 \\ 0 & 0 & a \end{bmatrix} \end{equation*}$$ Question: Considering the parameter value or values ​​obtained in item b), determine the associated space to its own value. Determine the algebraic and geometric multiplicity. Is A diagonalizable?

I' will solve this using: $$\det(A- \lambda I)$$: $$\begin{equation*} \det\begin{bmatrix} 0 & 0 & a \\ b & c & 10 \\ 0 & 0 & a \end{bmatrix} -\begin{bmatrix}\lambda&0&0\\ 0&\lambda&0\\ 0&0&\lambda\end{bmatrix} =\det\begin{pmatrix}-\lambda&0&a\\ b&c-\lambda&10\\ 0&0&a-\lambda\end{pmatrix} \end{equation*}$$

After we got the determinant we have: \begin{align*}(-\lambda)(c-\lambda)(a-\lambda)=0\end{align*}

And the solutions would be: \begin{align*} \lambda_1=0\\ \lambda_2=c\\ \lambda_3=a \end{align*}

The question b says: What value or values ​​should the parameters take for matrix A to have two real eigenvalues equal?

My answer would be: \begin{align*} a=c=0, b \in \mathbb{R} \end{align*}

And the following question says: Considering the parameter value or values ​​obtained in item b), determine the associated space to its own value(eigenvalue). Determine the algebraic and geometric multiplicity. Is A diagonalizable?

My professor said that i should give a and c some value, i didnt know about b so i give it a 0. \begin{align*} a=c=0, b \in \mathbb{R} \end{align*}

We calculate the eigenvalue of $$\lambda=0$$

We replace in: $$\begin{pmatrix}\lambda&0&a\\ b&c-\lambda&10\\ 0&0&a-\lambda\end{pmatrix}$$ $$\begin{pmatrix} 0 & 0 & a\\ b & c & 10\\ 0 & 0 & a \end{pmatrix}$$

We input the values of a,b,c we said before. $$\begin{pmatrix} 0 & 0 & a\\ b & c & 10\\ 0 & 0 & a \end{pmatrix}$$

And in row 3 we have that: \begin{align*} 10z=0\\ z=0 \end{align*}

We enter z value on the 2nd row \begin{align*} y+10z=0\\ y+10\cdot(0)=0\\ y=0 \end{align*}

The result would be \begin{align*} y=z=0 \longrightarrow \begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}x\\ 0\\ 0\end{pmatrix}=x\begin{pmatrix}1\\ 0\\ 0\end{pmatrix} \end{align*}

So we have that the eigenvalue is: \begin{align*} \lambda=0 \longrightarrow (1,0,0) \end{align*}

Is the process correct? If yes, then which would be the algebraic and geometric multiplicity?.

Besides that, should i calculate the same for $$\lambda=10$$ and $$\lambda=10$$?

• What did you mean by haunted proper space? – J. W. Tanner Mar 26 at 2:58
• @J.W.Tanner associated space sorry. – SWAT Mar 26 at 2:59
• What exactly do you mean by “$\lambda=0\rightarrow(1,0,0)$?” If $a=c=0$, then the matrix obviously has rank $1$ regardless of the value of $b$. This should tell you the dimension of the eigenspace associated with the only eigenvalue, which in turn tells you whether or not the matrix is diagonalizable. The eigenspace itself does depend on $b$, though. – amd Mar 26 at 4:59
• You write $a=c=0$, but then you don't use this when you write down the matrix. Your matrix should be $$\pmatrix{0&0&0\cr b&0&10\cr0&0&0\cr}$$ and now you need the eigenvalues and eigenvectors of this matrix. But the trouble is, you aren't justified in writing $a=c=0$. All you know is $a=c$. So you have to deal with the matrix $$\pmatrix{0&0&a\cr b&a&10\cr0&0&a\cr}$$ – Gerry Myerson Mar 26 at 8:29
• Any thoughts on my comment, Swat? – Gerry Myerson Mar 27 at 11:34