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Problem Statement

Assume $T^3(v) = 0$ for all $v$.

  • (a) Prove that $T$ is singular.

  • (b) Assume in addition that $T^2(v) \neq 0$. Prove that if $$\alpha v + \beta T(v) + \gamma T^2(v) = 0,$$ then $\alpha = \beta = \gamma = 0$.

My Attempt

(a) I really have no idea where to start on this. I do know that if T is singular then that means there does not exist a B such that $TB = BT = I$. However, I don't know where to go from this.

(b) I'm tempted to just continue taking transformations of both sides until I get $0 = 0$ but I don't see how that helps me show that $\alpha = \beta = \gamma = I$.

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    If $T$ is nonsingular, prove the same is true for $T^3$. Can you do this? – 2017-02-12
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    @PedroTamaroff I believe so. I want to say that I would start with letting $T$ be nonsingular, then $T^3 = (B^{-1} \cdot I)^3 = (B^{-1})^3 \cdot I^3 = (B^{-1})^3 = ((T^{-1})^{-1})^3 = T^3$. I feel that is the relation I would need to use. – 2017-02-12
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    For the statement to be true, the vector space on which T acts has to be of nonzero dimension. – 2017-02-12
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    You probably mean that $T^3(v)=0$ for *some* $v$? – 2017-02-12
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    This is what my professor printed on the paper. – 2017-02-12

3 Answers 3

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a) If $T^3(v) = 0$ for all $v$, then $T^3$ must be the zero matrix. Thus we have $T^3 = 0$. Taking the determinate of both sides, we have $0 = det(0) = det(T^3) = det(T*T*T) = (det(T))^3$. If $det(T)^3 = 0$ we must have that $det(T) = 0$.

b) If we multiply both sides by $T^2$, we have $\alpha T^2(v) + \beta T^3(v) + \gamma T*(T^3)(v) = 0$ using the fact that $T^3(v) = 0$ and $T^2(v) \neq 0$, we have $\alpha T^2(v) = 0$ and hence, we must have that $\alpha = 0$. Now we have that $\beta T(v) + \gamma T^2(v) = 0$. Multiplying both sides by $T$ and using a similar argument will give that $\beta = 0$. Finally, with this, we simply have $\gamma T^2(v) = 0$ and hence, $\gamma = 0$.

Hope this helps!

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    How can you make the assumption that $T$ must be the zero matrix if $T^3(v) = 0$ for all of $v$? – 2017-02-12
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    $T^3(v) = 0$ for all $v$, means that the null space of $T^3$ must be full (i.e it must span all of whatever space $v$ comes from). So, by the rank-nullity theorem, $T^3$ must be rank 0. The only rank 0 matrix is the zero matrix. – 2017-02-13
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Suppose there is some $v\neq 0$ such that $T^3v=0$. If $T^2v$ is nonzero, then $Tw=0$ for $w=T^2v$, so $T$ is singular. If $T^2v$ is zero, then either $Tv=w'$ is nonzero, so $Tw'=0$ and $T$ is singular, or $Tv=0$ is zero, so $T$ is singular. This is, in fact, an inductive proof that if $T^n$ is singular for some $n$, then $T$ is also singular.

Suppose now that $\alpha v + \beta T(v) + \gamma T^2(v) = 0$, with $T^2v$ nonzero. Apply $T^2$ to get

$$\alpha T^2(v) + \beta T^3(v) + \gamma T^4(v) = 0.$$

The terms involving $\beta$ and $\gamma$ are zero, and you get $\alpha T^2(v) =0$, which gives $\alpha=0$. Continue to do this, and try to provide a similar inductive proof that if $T^n(v)=0$ but $T^{n-1}(v)$ is nonzero, then...?

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I assume we're working in a vector space $V$ over some field $F$ (you can take $F=\mathbb{R}$ if you want) which is not $\{0\},$ and $T$ is a linear transformation $V\to V$ (if you prefer, $T$ is a square matrix with entries in $F$). The identity is $I.$

$\text{(a)}$ There is a slick way and a more "hands-on" way.

Slick way: If $T$ is non-singular then $T$ is a bijection $V\to V,$ so that in particular $\operatorname{im}{(T)}=T(V)=V.$ Now $$\operatorname{im}{(T^{2})} = T^{2}(V) = T(T(V)) = T(V) = V.$$ Similarly, $\operatorname{im}{(T^{3})} = V.$ But if $T^{3}(v)=0$ for all $v\in V,$ it must follow that $V=\{0\},$ a contradiction. Hence, if $T^{3}(v)=0$ for all $v\in V,$ then $T$ is not non-singular, and hence $T$ is singular.

Hands-on way: Suppose $T$ is non-singular. Then there is an inverse $T^{-1}$ with $TT^{-1}=T^{-1}T=I.$ We are told that $$0=T^{3}(v)$$ for all $v\in V.$ Then, applying $T^{-1}$ to both sides (if you prefer, left-multiplying by $T^{-1}$ on both sides) gives $$0 = T^{-1}(0) = T^{-1}T^{3}(v) = T^{2}(v).$$ But now we can do the same thing, to get $T(v)=0$ and hence $v=0.$ Since we assumed $V\neq\{0\},$ we know this is not true for all $v\in V,$ so we have a contradiction; hence $T$ is not non-singular, so $T$ is singular.

$\text{(b)}$ Here I assume $v\in V$ is supposed to be fixed (the question does not make sense otherwise). Since $T$ is a linear transformation (or square matrix) and $T(T(v))\neq0,$ it follows that $T(v)\neq0,$ and by the same reasoning $v\neq0.$ We have an equation \begin{align} \alpha v+\beta T(v)+\gamma T^{2}(v)=0.\tag{1} \end{align} Apply $T$ to both sides of $(1)$ (if you prefer, left-multiply both sides by $T$). Since $T$ is linear and $T^{3}(v)=0,$ this gives \begin{align} \alpha T(v)+\beta T^{2}(v)=0.\tag{2} \end{align} Now, do the same thing to $(2)$ to get \begin{align} \alpha T^{2}(v)=0.\tag{3} \end{align} Now we know $T^{2}(v)\neq0,$ so we deduce from $(3)$ that $\alpha=0.$ Hence $(2)$ becomes $$\beta T^{2}(v)=0,$$ and similarly we deduce that $\beta=0.$ Finally the same reasoning applied to $(1)$ gives $\gamma=0$ and we are done.