# Prove that the operator induced by $T$ on the quotient space $V/\ker[T-5I]$ has all eigenvalues $=0$

The question is as follows : A linear operator T on a complex vector space V has a characteristic polynomial $x^3(x-5)^2$ and minimal polynomial $x^2(x-5)^2$ .

Now, i need to prove that the operator induced by T on the quotient space $V/\ker[T-5I]$ has all eigen values = $0$.

$Attempt$: The matrix from the given information can be deduced as follows :

\begin{pmatrix} 0 &1 &0 &0 &0\\ 0 &0 &0 &0 &0\\ 0 &0 &0 &0 &0\\ 0 &0 &0 &5 &0\\ 0 &0 &0 &0 &5 \end{pmatrix}

$$W = \ker(T-5I) = \operatorname{span} [(0,0,0,1,0)^T, (0,0,0,0,1)^T ].$$

The basis of the quotient is given by: $[ W+(1,0,0,0,0)^T, W+(0,1,0,0,0)^T , W+(0,0,1,0,0)^T]$.

Now, I don't understand specifically much by the phrase operator induced by T on the quotient space spanned by the above basis , but , i think it means T acting on a vector in the quotient space i.e. $T [ \alpha_1 (W + [1,0,0,0,0]^T ) + \alpha_2 (W + [0,1,0,0,0]^T + \alpha_3 (W + [0,0,1,0,0]^T ]$

\begin{align} & = T [W + \alpha_1 [1,0,0,0,0]^T + \alpha_2 [0,1,0,0,0]^T + \alpha_3 [0,0,1,0,0]^T ] \\ & = Tw + \alpha_1 .0 + \alpha_2 [1,0,0,0,0]^T + \alpha_3 .0 \end{align} $w$ is a vector which belongs to $W$

$$= T [ \alpha_4 [(0,0,0,1,0)^T + \alpha_5 (0,0,0,0,1)^T ] = 5\alpha_4 [(0,0,0,1,0)^T + 5\alpha_5 (0,0,0,0,1)^T$$

$$=(0,0,0,5\alpha_4, 5\alpha_5)^T$$

and the eigenvalues don't come as $0$ as desired in the question statement. Where am i going wrong in understanding the question? Thanks.

• Your way of writing subscripts and some of your other ways of writing MathJax code are bizarre. Please look at my edits. You shouldn't keep alternating in and out of MathJax so many times. – Michael Hardy Dec 5 '13 at 19:54
• Alright. Will take care of it. – MathMan Dec 5 '13 at 20:10

## 2 Answers

Your definitions of quotient operator and its eigenvalue are wrong, hence the confusion.

First of all, look at what you think of as the quotient operator over $W$ $$T(W+α_1[1,0,0,0,0]^T+α_2[0,1,0,0,0]^T+α_3[0,0,1,0,0]^T)$$ - let's call this $T(W+v)$, being equal to this $$Tw+α_1.0+α_2[1,0,0,0,0]T+α_3.0 = Tw + Tv\quad(1)$$
You have defined a map from an (affine) subset to a single vector: $$T(v+W) = Tv+Tw,\space w\in W$$ which means each set $v+W$ could be mapped to multiple outputs corresponding to multiple $w$ by the same map $T$. Thus this definition doesn't make sense unless you specify how to pick $w$ through $T$.

The better definition of quotient operator is: $$T|_W: V/W\to V/W$$ $$T|_W(v+W) = Tv+W$$ which maps from an affine subset to another affine subset in the same vector space (hence the name operator). You can verify that this mapping actually makes sense.

I hope you are familiar with algebraic operations on quotient space: $$(v+W)+(u+W)=v+u+W$$ and $$k(v+W) =kv +W$$ Note that this means the additive identity on $V/W$ is $W$

And then we can define eigenvalue of quotient operator $T|_W$:

$\lambda$ is an eigenvalue of $T|_W$ $\iff$ $\exists v \notin W: T|_W(v+W)=\lambda v+W$

Using these definitions on $(1)$ you will arrive at the desired result.

The quotient vector space has dimension $3$, and is spanned by the classes of the first three basis vectors. This you have computed above. In other words, the linear map on $W$ with respect to this basis has the matrix $$\begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0\cr 0 & 0 & 0 \end{pmatrix}.$$ This matrix has only $0$ as an eigenvalue, with multiplicity $3$.

• if i denote the first three basis vectors as a,b,c ; then the quotient space is spanned by the basis { W+a , W+b, W+c }. Then, why are we just writing incorporating in the matrix which you wrote T(a), T(b) and T(c), i mean why no role of T(w), where w belongs to W [ as transformation matrix incorporates T acting on individual basis elements of the vector(quotient) space ], if you get what i mean. – MathMan Dec 5 '13 at 20:16
• @@Dietrich Burde) Similarly the matrix of $V/ker(T)$ is $5I_{2\times 2}$. Am I right ? – Empty Dec 10 '15 at 18:39