# A condition for an operator $T$ be diagonalizable

Let $V$ be $n$-dimensional and $T: V \rightarrow V$ have characteristic polynomial

$$(x-c_{1})^{d_{1}}...(x-c_{k})^{d_{k}} \quad\text{where}\quad d_{1} + ...+ d_{k} = n$$ and the $c_i$ are pairwise distinct. My book gives the following statement:

$T$ is diagonalizable if and only if $\dim\ker(T-c_{i}I)=d_i$ for $i=1,\ldots,k$.

Why does $\ker(T-c_{i}I)$ then have dimension $d_{i}$? And how do I prove the converse?

• i know that $V$ have a basis where at least one vector are in the kernel of $T-c_{i}I$ but why there is exact $d_{i}$ elements of the basis there? – Eduardo Silva Jan 24 '16 at 3:12
• Please be so kind as to spell out what you mean by "the converse". Explanation: usually one uses "converse" for a statement of the form "if A then B", for which the converse reads "if B then A". There is one statement of that form in the question, namely the first paragraph. By this reasoning the part about $\dim\ker(T-c_{i}I)=d_i$ has no place in the converse; however then the converse is not true (as I mentioned in my answer). It might make more sense if "the converse" would contain something about $\dim\ker(T-c_{i}I)=d_i$; if that is the intention, please write out the intended converse. – Marc van Leeuwen Jan 25 '16 at 10:05
• sorry, the question from the book was "T is diagonalizable if and only if $dim Ker(T-c_{i}T) = d_{i}$ with the polynomial above. – Eduardo Silva Jan 25 '16 at 21:25
• I've edited the question to say what (I suppose) you meant. – Marc van Leeuwen Jan 26 '16 at 10:01

The main ingredient for a proof is that $$1\le\dim\ker(T-c_iI)\le d_i$$ (which will be discussed later).

Since $T$ is diagonalizable, there is a basis $\mathscr{B}$ consisting of eigenvectors of $T$. Write $$\mathscr{B}=\mathscr{B}_1\cup\mathscr{B}_2\cup\dots\cup\mathscr{B}_k$$ where each element of $\mathscr{B}_i$ is an eigenvector relative to $c_i$. Now set $$m_i=\dim\ker(T-c_iI),\qquad m'_i=|\mathscr{B}_i|$$ We have, for $i=1,2,\dots,k$, $$m'_i\le m_i\le d_i$$ The first inequality is because $\mathscr{B}_i$ is a linearly independent set in $\ker(T-c_iI)$.

Summing up the inequalities, we get $$n=\sum_{i=1}^k m'_i\le\sum_{i=1}^k m_i\le\sum_{i=1}^k d_i=n$$ and therefore, for each $i$, $$m'_i=m_i=d_i$$

Proof of the main ingredient. Suppose $c$ is an eigenvalue of $T$ and that $(c-x)^d$ divides the characteristic polynomial $\det(T-xI)$, but $(c-x)^{d+1}$ does not divide it.

Let $\{v_1,v_2,\dots,v_m\}$ be a basis of $\ker(T-cI)$. Extend it to a basis $\mathscr{B}=\{v_1,\dots,v_m,v_{m+1},\dots,v_n\}$ of $V$. Then the matrix of $T$ relative to the basis $\mathscr{B}$ has the form $$\begin{bmatrix} cI_m & B \\ 0 & C \end{bmatrix}$$ because, by assumption, $T(v_i)=cv_i$, for $i=1,2,\dots,m$. Then the characteristic polynomial is $$\det\begin{bmatrix} cI_m-xI_m & B \\ 0 & C-xI_{n-m} \end{bmatrix} =(c-x)^m\det(C-xI_{n-m})$$ which proves that $m\le d$, because $(c-x)^m$ divides the characteristic polynomial.

For the converse, you need to be in an algebraically closed field, so certainly the characteristic polynomial can be factorized as $$(c_1-x)^{d_1}(c_2-x)^{d_2}\dots(c_k-x)^{d_k}$$ or, at least, you need that such a factorization exists for the specific $T$.

Assume that, for each $i=1,2,\dots,k$, we have $d_i=\dim\ker(T-c_iI)$. Take a basis for each eigenspace $\ker(T-c_iI)$ and put them together. This gives a basis for $V$, because the eigenspaces are independent.

One can compute the characteristic polynomial of $T$ using a matrix for $T$ with respect to any basis. Assuming that $T$ is diagonalisable, we can use a diagonal matrix, and then clearly the characteristic polynomial is the product of factors $x-c$ where $c$ runs over the $n$ diagonal coefficients of the matrix. If $k$ is the number of distinct diagonal entries (eigenvalues) and the eigenvalue $c_i$ occurs $d_i$ times as diagonal entry, for $1=1,\ldots,k$, then the characteristic polynomial is $(x-c_1)^{d_1}\ldots(x-c_k)^{d_k}$ (as the question states) and clearly $d_1+\cdots+ d_k = n$. The matrix of $T-c_iI$ on the basis of diagonalisation is a diagonal matrix with $d_i$ diagonal entries equal to $0$, so its kernel is the span of the basis vectors corresponding to those entries $0$, which clearly has dimension$~d_i$.

So the "direct" part of the question is quite immediate. However the converse that the question asks about is not true (having such a characteristic polynomial does not imply that $T$ is diagonalisable), and indeed there exists no condition for $T$ to be diagonalisable in terms of the characteristic polynomial alone, as the title of the question would suggest. This can be seen by considering an upper triangular matrix with equal entries$~c$ on the diagonal: its characteristic polynomial is always $(x-c)^n$, but the matrix is diagonalisable only if it is already diagonal (i.e., all coefficients above the diagonal happen to be $0$).

(After modification of the question:) Now that it has become clear that the converse the question was asking about was in fact: (given that characteristic polynomial) "if $\dim\ker(T-c_i)=d_i$ for all $i$ then $T$ is diagonalisable", let me answer how one shows that. The key point is the following general property:

For any linear operator $T$ and any distinct scalars $c_1,\ldots,c_k$, the subspaces $\ker(T-c_i)$ always for a direct sum.

These subspaces are eigenspaces whenever they have nonzero dimension, as is the case here (because by hypothesis the $c_i$ are roots of the characteristic polynomial of $T$). Assuming the property, it tells us that the sum of those eigenspaces has dimension $d_1+\cdots+d_k=n$ (because the dimension of a sum of subspaces is the sum of their dimensions if (and only if) the sum is a directs sum), whence it is all of$~V$, and this means (for me by definition) that $T$ is diagonalisable. (One can explicitly get a basis consisting of eigenvectors by taking the union of bases of the subspaces $\ker(T-c_i)$; again the essential point is that their sum is direct, so that this gives a basis of the sum, in other words of$~V$.)

So why do we have that general property the sums of distinct eigenspaces are always direct? This can be proved by induction on the number of values$~c_i$, the case of one value being trivial (the "sum" of a single subspace is always direct). So suppose one has a direct sum $S$ of subspaces $\ker(T-c_iI)$ for $i=1,\ldots,k-1$, and consider the additional subspace $W=\ker(T-c_kI)$. To show that the sum remains direct upon including $W$, amounts to showing that $S\cap W=\{0\}$. Taking $v\in S\cap W$, one can on one hand decompose $v=v_1+\cdots+v_{k-1}$ as a sum of its components in the direct sum $S$, and since the summands are eigenspaces for $c_1,\ldots,c_{k-1}$ respectively, one has $Tv=c_1v_1+\cdots+c_{k-1}v_{k-1}$. On the other hand, since $v\in W$ one has $Tv=c_kv=c_kv_1+\cdots+c_kv_{k-1}$. Uniqueness of the decomposition in a direct sum gives that $c_1v_1=c_kv_1,\ldots,c_{k-1}v_{k-1}=c_kv_{k-1}$, and since $c_k$ differs from each of $c_1,\ldots,c_{k-1}$, it follows that $0=v_1=\cdots=v_{k-1}$, so $v=0$ as desired.

• But by the Cayley-Hamilton Theorem $f(T) = (T-c_{1}I)^{d_{1}}...(T-c_{k}I)^{d_{k}} = 0$, meaning that there is a basis $\{v_{i}\}$ where $f(T)v_{i} = 0$, therefore, all $v_{i}$ are eigenvalues and thats the definition of diaonalizable – Eduardo Silva Jan 24 '16 at 13:50
• No (at least what you suggest is wrong). The $v_i$ (indeed any nonzero vectors) are eigenvalues of $f[T]=0$, but not eigenvectors of $T$. If as you you suggest eigenvectors of $f[T]$ were automatically eigenvalues of $T$, that would mean that by the C-H theorem all matrices are diagonalisable, but that of course is not true. – Marc van Leeuwen Jan 24 '16 at 16:47
• I think the point is to use the condition on the dimension of the kernel as part of the converse. (Though the question is a little unclear to me.) I don't see a way to avoid using the Jordan Canonical form. Do you? – Lorenzo Jan 24 '16 at 17:27

Hint (to why T diagonalizable implies the characteristic polynomial is that way, and how to compute the kernel of $T - c_i I$): Computing the kernel doesn't depend on the basis, and the characteristic polynomial also is invariant of basis. Choose the basis in which $T$ is diagonalized, and compute the characteristic polynomial and the kernel of $T - c_i I$.

For the converse, it's a little harder.

It is not true that the characteristic polynomial having that form implies that it can be diagonalized. You can find an upper triangular 2x2 matrix which is of the form $I + N$, where $N^2 = 0$. The nilpotent piece of the counter example (N) is precisely the reason why you need this assumption about the dimension of the kernel of $T - c_i I$.

The precise statement:

Let $k$ be a field. If the characteristic polynomial $f$ of $T : k^n \to k^n$ splits (over $k$) as $f (x) = \Pi (x - c_i)^{n_i}$, where $\Sigma n_i = n$, and for each $i$, $dim$ $Ker(T - c_i I) = n_i$, then $T$ is diagonalizable (over $k$).

Proof sketch (via giant hammer): Since all the eigenvalues lie in the field $k$, there is a basis in which $T$ has a Jordan canonical form. If any Jordan blocks $J_{\lambda}$ are not of size $1 \times 1$ (i.e. there are nonzero entries above the diagonal), then $ker(T - \lambda I)$ is too small.

• dim $ker(T−c_iI)=n_i$? – Eduardo Silva Jan 24 '16 at 17:22
• @EduardoSilva What is your question /comment? I don't understand. – Lorenzo Jan 24 '16 at 17:24
• at this passage of yours "and for each $i$, $Ker(T-c_{i}I) = n_{i}$" , I thought that you was trying to say $dim Ker(T-c_{i}I) = n_{i}$ – Eduardo Silva Jan 24 '16 at 17:31
• @EduardoSilva Yes, that's correct. – Lorenzo Jan 24 '16 at 17:31