# Show the following equivalence regarding the diagonalization of matrices.

Let $$T$$ be a linear operator on a finite dimensional vector space $$V$$. Let $$\lambda_1 , \lambda_2 , \cdots , \lambda_k$$ be the distinct eigen values of $$T$$. Let $$W_1 , W_2 , \cdots , W_k$$ be the eigen spaces corresponding to the distinct eigen values $$\lambda_1 , \lambda_2 , \cdots , \lambda_k$$ respectively. Then the following are equivalent $$:$$

$$(1)$$ $$T$$ is diagonalizable.

$$(2)$$ $$V=W_1+W_2+ \cdots + W_k$$.

My attempt $$:$$

I have assumed two lemmas which are as follows :

Lemma-1 :

Let $$T$$ be a linear operator on a finite dimensional vector space $$V$$.Then if $$T$$ is digonalizable then it's eigen values are all regular.

This lemma can be proved easily by observing that if $$T$$ is diagonalizable then it's characteristics polynomial splits completely into linear factors and by looking at the null space of $$T-\lambda I$$ for each eigen value $$\lambda$$ of $$T$$ whose diagonal matrix representation contains exactly $$d$$ zeros along the main diagonal if the algebraic multiplicity of $$\lambda$$ is $$d$$.

Lemma-2 :

Let $$T$$ be a linear operator on a finite dimensional vector space $$V$$ and let $$f$$ be a polynomial over the field of $$V$$ in one variable $$t$$. If $$x$$ be the eigen vector of $$T$$ corresponding to the eigen value $$\lambda$$ then $$f(T)x=f(\lambda)x$$.

Which can also be proved easily by observing that $$T^n x =\lambda^n x$$ for each $$n \in \mathbb N$$.

Lets now prove $$(1) \implies (2)$$.

Let $$W=W_1+W_2+ \cdots +W_k$$.

Claim $$:$$ $$\dim W = \dim W_1 + \dim W_2 + \cdots + \dim W_k$$.

Let $$B_j = \{u^{j1},u^{j2},\cdots ,u^{jd_j} \}$$ be the basis of $$W_j$$ for $$1 \leq j \leq k$$. Then it follows easily that $$W$$ is spanned by $$B_1 \cup B_2 \cup \cdots \cup B_k$$.To prove that $$\cup B_j$$ is linearly independent let us consider the equation

$$c_{11}u^{11} + \cdots + c_{1d_1}u^{1d_1} + c_{21}u^{21}+\cdots+c_{2d_2}u^{2d_2}+\cdots+c_{k1}u^{k1}+\cdots+c_{kd_k}u^{kd_k}=0$$.

Then for any polynomial $$f$$ applying $$f(T)$$ to both sides of the above equation and by applying lemma-2 we have

$$c_{11}f(\lambda_1)u^{11} + \cdots + c_{1d_1}f(\lambda_1)u^{1d_1} + c_{21}f(\lambda_2)u^{21}+\cdots+c_{2d_2}f(\lambda_2)u^{2d_2}+\cdots+c_{k1}f(\lambda_k)u^{k1}+\cdots+c_{kd_k}f(\lambda_k)u^{kd_k}=0$$.

In particular if we take the polynomial $$f$$ to be $$f_j$$ where $$f_j$$ is defined as

$$f_j(x)=(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_{j-1})(x-\lambda_{j+1})\cdots(x-\lambda_k)$$ for $$1 < j < k$$; $$f_j(x)=(x-\lambda_2)(x-\lambda_3)\cdots(x-\lambda_k)$$ for $$j=1$$ and $$f_j(x)=(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_{k-1})$$ for $$j=k$$ then we have

$$c_{j1}=c_{j2}=\cdots=c_{jd_j}=0$$ for $$1 \leq j \leq k$$.

This proves that $$\cup B_j$$ is linearly independent and hence a basis for $$W$$ and therefore $$\dim W=\dim W_1+\dim W_2+\cdots+\dim W_k$$ which proves our claim.

Now since $$T$$ is diagonalizable so by using lemma-1 we have

$$\dim W_j =$$ geometric multiplicity of $$\lambda_j =$$ algebraic multiplicity of $$\lambda_j$$.

Since $$T$$ is diagonalizable it's characteristics polynomial splits completely into linear factors and hence the sum of the algebraic multiplicities of the eigen values of $$T$$ should add upto $$\dim V$$. Hence $$\dim W = \dim V$$. But $$W$$ is a subspace of $$V$$. So it follows that $$W=V$$ and cosequently $$V=W_1+W_2+\cdots+W_k$$ and we are done.

$$(2) \implies (1)$$ $$:$$

If $$V=W_1+W_2+\cdots+W_k$$. Then as in the proof above $$\cup B_j$$ is linearly independent. Let $$|B_j|=d_j$$ for $$1 \leq j \leq k$$ and let $$\dim V=n$$. Then from the above proof it follows that $$n=d_1+d_2+\cdots+d_n$$. Let $$B=\cup B_j$$. Then $$B$$ is a linearly independent set of $$n$$ vectors from $$V$$ and since $$\dim V=n$$ so $$B$$ is a basis for $$V$$. So we have a basis $$B$$ for $$V$$ each vector of which is an eigen vector of $$T$$ and consequently the matrix representation of $$T$$ relative to that ordered basis $$B$$ is a diagonal matrix. Therefore $$T$$ is diagonalizable.

Hence the result follows.

Now my question is : Is my reasoning above ok? Though I am not satisfied with my proof. The main reason behind it that the proof is too large and hence laborious. Is there any easy shorter way of proving this result? Please help me by proving this if there is any.

Thank you in advance.

• What is the meaning of "regularity of eigenvalues " in lemma 1? Oct 5, 2017 at 7:03
• If algebraic and geometric multiplicity of an eigen value is same then the eigen value is said to be regular.
– user251057
Oct 5, 2017 at 7:06
• :Your proof is perfectly fine!! Oct 5, 2017 at 7:40
• :From where you get this proof? Oct 5, 2017 at 8:05
• I have proved it myself by some hints given by my teacher.
– user251057
Oct 5, 2017 at 8:19

Unless I'm missing something, I think the proof should be a lot shorter, the result should be almost immediate. Your proof for $(2) \implies (1)$ is nice, here's an alternative for the other direction.
$(1) \implies (2)$. Since $T$ is diagonalisable, there is a basis $v_1, \ldots, v_n$ of $V$ (where $n = \dim V$) such that $T v_i = a_i v_i$ for some scalars $a_1, \ldots, a_n$. These scalars must be eigenvalues, and setting $W_j = \mathrm{span}\{v_i \mid a_i = \lambda_j\}$ gives that $V = W_1 + \ldots + W_n$, since any basis vector $v_i$ must lie in one of the $W_j$.