# Prove Φ is a linear transformation

Suppose $\dim U_1 + \dim U_2 + \dots + \dim U_m = \dim V$. Show that there exist an isomorphism $\Phi : V \to W$ such that $\Phi (T(v)) = \Psi (\Phi (v))$ for every $v \in V$.

$T:V \rightarrow V$ is a linear operator on a finite-dimensional vector space V. $U_i$ is the eigenspace of T. $Ψ:W\rightarrow W$ is a linear operator on W. W is the vector space $U_1 ⊕ U_2 ⊕ ... ⊕ U_m$.

I am solving this question, and I am up to this point where I need to first show $Φ : V\rightarrow W$ is a linear transformation.

To prove vector addition, I set up $Φ(T(v_1+v_2)$, and apply the definition of Φ so that I have $Ψ(Φ(v_1+v_2)$, but then at this point I realize I need to assume it is a linear transformation so I can say $Φ(v_1+v_2)=Φ(v_1)+Φ(v_2)$. After this assumption, I can say the equation $=Ψ(Φ(v_1)+Ψ(Φ(v_2)=Φ(T(v_1))+Φ(T(v_2))$

Then how can I prove this is a linear transformation?

• There is context missing. What are the $U_i$? – darij grinberg Dec 4 '16 at 5:51
• @darijgrinberg sorry, I just edited it in the question. I didn't think it is related to the prove linear transformation part so I didn't add it. – Vanya Dec 4 '16 at 5:53
• OK, then what is $\Psi$? – darij grinberg Dec 4 '16 at 5:54
• @darijgrinberg sorry that's my bad. Should be all now. The Ψ linear operator is actually given, but it is not related to the linear transformation. – Vanya Dec 4 '16 at 5:59
• Okay, but if $\Psi$ is just any random endomorphism of $W$, then this is completely wrong. Probably some specific map is required (my guess is the map that acts on each $U_i$ by multiplication by the eigenvalue). Either way, I am afraid your understanding seems to end significantly beneath the level of the problem; for example, you cannot "set up" a map $\Phi$ by requiring $\Phi(T(v)) = \Psi(\Phi(v))$, because that's not an explicit equation defining a value of anything under $\Phi$ (after all, $\Phi$ appears on both sides!). Instead, you need to construct it in a different way. – darij grinberg Dec 4 '16 at 6:04

## 2 Answers

From what I can tell, you aren't even given a defintion of $\Phi: V \rightarrow W$. I believe what you are supposed to do is define the function yourself, then show it is a linear transformation.

I am guessing $\Psi: W \rightarrow W$ is the linear transformation $(u_1, ... , u_m) \mapsto (\lambda_1 u_1, ... , \lambda_m u_m)$.

In that case, it is easier to define a linear transformation of $W = U_1 \oplus \cdots \oplus U_m$ into $V$, not the other way around. By a standard result, namely the linear independence of distinct eigenspaces, the natural map $(u_1, ... , u_m) \mapsto u_1 + \cdots + u_m$ is an injective linear transformation $S: W \rightarrow V$.

The image of $S$ in $V$ is a subspace whose dimension is $\textrm{Dim } W = \sum\limits_{i=1}^m \textrm{Dim } U_i = \textrm{Dim } V$. Since the image of $S$ is contained in $V$, and they have the same dimension, they must be equal. Hence $S$ is a linear transformation which is a vector space isomorphism (linear transformation and a bijection).

You can then define $\Phi$ to be the inverse function of $S$. It is a linear transformation, because $S$ is.

Now if $v$ is in $V$, let $\Phi(v) = (u_1, ... , u_m)$ for some $u_i \in U_i$, so that $v = u_1 + \cdots + u_m$.

Then $$\Phi \circ T(v)= \Phi(\lambda_1u_1 + \cdots + \lambda_m u_m) = (\lambda_1u_1, ... , \lambda_m u_m) = \Psi(u_1, ... , u_m) = \Psi \circ \Phi(v)$$

• Thanks for you answer! I only have one question. How do we know $T(v)=λ_1u_1+λ_2u_2+...+λ_mu_m$? – Vanya Dec 4 '16 at 18:16
• Is it possible to show W is a subspace of V? If so, I can then say W = V because they have same dimension, and therefore $T(v)=λ_1u_1+λ_2u_2+...+λ_mu_m$? – Vanya Dec 4 '16 at 18:21
• For your first question, (I assume) you are given that $U_i$ is the subspace of $V$ consisting of all those $u \in V$ such that $T(u) = \lambda_i u$. – D_S Dec 4 '16 at 19:13
• For your second question, that depends on what you mean by $W = U_1 \oplus \cdots \oplus U_m$ (direct sum). In my answer, I meant it as the external direct sum, which is to say that $W$ is literally the cartesian product $$U_1 \times \cdots \times U_m$$ which is given the structure of a vector space. So $W$ is not literally a subspace of $V$; however, it is isomorphic to a subspace of $V$ via the linear transformation $S$. – D_S Dec 4 '16 at 19:16
• They make a lot of sense! Thank you! – Vanya Dec 4 '16 at 19:25

Lemma: Let $T$ be a linear operator on $V$. Let $u_1, ... , u_t$ be nonzero eigenvectors of distinct eigenvalues $\lambda_1, ... , \lambda_t$ for $T$. Then $u_1, ... , u_t$ are linearly independent.

Proof: By induction on $t$. If $t = 1$, this is immediate. More generally, assume $c_1u_1 + \cdots + c_tu_t = 0$ for scalars $c_i$. One of these scalars, say $c_1$, is not zero. Multiplying this equation by the scalar $\lambda_1$, we get

$$\lambda_1c_1 u_1 + \cdots + \lambda_1 c_t u_t = 0$$

Alternatively, we can apply $T$ to get

$$\lambda_1c_1u_1 + \cdots + \lambda_t c_tu_t = 0$$

Subtracting these two equations gives

$$(\lambda_2 - \lambda_1)c_2u_2 + \cdots + (\lambda_t - \lambda_t)c_tu_t = 0$$

By induction, $u_2, ... , u_t$ are linearly independent, so all the scalars $(\lambda_i - \lambda_1)c_i$ ($i =2, ... , t$) are zero. But $\lambda_i - \lambda_1 \neq 0$, so $c_i = 0$. $\blacksquare$

Proposition: Let $T$ be a linear operator on $V$, with eigenvalues $\lambda_1, ... , \lambda_m$. Let $U_i$ be the eigenspace of $\lambda_i$, and let $W$ be the (external) direct sum $U_1 \oplus \cdots \oplus U_m$. Define a function $S: W \rightarrow V$ by

$$S(u_1, ... , u_m) = u_1 + \cdots + u_m$$

Then $S$ is a linear transformation, and it is injective.

Proof: Let $w = (u_1, ... , u_m), w' = (u_1', ... , u_m')$ be elements of $W$, and let $c$ be a scalar. Then

$$S(w+cw') = S(u_1 + cu_1', ... , u_m + cu_m')= u_1 + cu_1' + \cdots + u_m + cu_m'$$

$$= u_1 + \cdots + u_m + cu_1' + \cdots + cu_m' = S(w) + S(cw')$$

so $S$ is a linear transformation.

To show that $S$ is injective, it suffices to show that if $w = (u_1, ... , u_m) \in W$, and $S(w) = 0$, then $w = (0, ... , 0)$. So by hypothesis,

$$u_1 + \cdots + u_m = 0$$

But nonzero eigenvectors of distinct eigenvalues are linearly independent by the lemma, so we must have $u_1 = \cdots = u_m = 0$.

• Thanks for your explanation! I completely understood your answer to the question and the Lemma, but I wonder how in the world you came up the solution tricky like that.. – Vanya Dec 4 '16 at 20:16