2
$\begingroup$

I am following a book where the "diagonalizability" has been introduced as follows:

Consider a basis formed by a linearly independent set of eigen vectors $\{v_1,v_2,\dots,v_n\}$. Then it is claimed that with respect to this basis, the matrix $A$ is diagonal.

I am confused at the word "basis" here. In some other books it is said that a matrix $A$ is called diagonalizable if there exists matrix $P$ such that $P^{-1}AP$ is a diagonal matrix.

I don't think that diagonalizability has anything to do with basis. It just happened that the set of eigen vectors form a basis of $\Bbb R^n$. Also I don't think having a set of $n$ linearly independent eigen vectors is a necessary condition for a matrix to be diagonalizable.

$\endgroup$
4
  • $\begingroup$ The motivation of diagonalising comes from basis. Both the above definations are equivalent. Linear transformations $A$ and $B$ are said to be equivalent iff there exists an invertible matrix $P$ such that $P^{-1}AP=B$. Thats because that operations is just writing $A$ in another basis. $\endgroup$
    – Grobber
    Aug 25 '13 at 12:41
  • $\begingroup$ Try reading Linear Transformations once. That would make it clear. $\endgroup$
    – Grobber
    Aug 25 '13 at 12:42
  • $\begingroup$ @Grobber What is that basis corresponding to the final diagonal matrix ? $\endgroup$ Aug 25 '13 at 13:34
  • $\begingroup$ The basis desired are the column vectors of $P$. $\endgroup$
    – Grobber
    Aug 25 '13 at 16:04
3
$\begingroup$

A complex matrix $A\in\mathbb{C}^{n\times n}$ is diagonalizable iff its eigenvectors form a basis of $\mathbb{C}^n$ (note that even real matrices can have complex eigenvectors). That is there exists a nonsingular matrix $P$ such that $P^{-1}AP=D$, where $D$ is diagonal containing the eigenvalues of $A$ with corresponding eigenvectors being the columns of $P$. Also, $P^{-1}AP=D$ means that $AP=PD$, that is, $A$ acts in the basis formed by columns $P$ as diagonal matrix ($D$).

Some notes:

Not all matrices are diagonalizable. For example nontrivial Jordan blocks: $$ A = \begin{bmatrix} 1 & 1 & & & & \\ & 1 & 1 & & & \\ & & \ddots & \ddots & & \\ & & & 1 & 1 \\ & & & & 1 \end{bmatrix}\in\mathbb{R}^{n\times n} $$ has 1 eigenvalue of multiplicity $n$ but with only one (linearly independent) eigenvector.

A matrix with distinct eigenvalues is diagonalizable.

$\endgroup$
2
  • $\begingroup$ If $A$ is diagonalizable i.e. $AP=PD$ then it is clear that the set $P$ contains all the eigen vectors of $A$ and the diagonal entries of $D$ are eigen values of $A$. How to prove that the eigen values are distinct from here (which implies that eigen vectors are independent). $\endgroup$ Aug 25 '13 at 14:40
  • 1
    $\begingroup$ I'm not sure what the question is about. The diagonalizability does not imply that the matrix has distinct eigenvalues (like an identity for example). The reverse is true though and can be proved directly using the definitions of the terms involved (a simplified hint: two distinct eigenvalues cannot share the same eigenvector). $\endgroup$ Aug 26 '13 at 13:30
2
$\begingroup$

The two definitions are equivalent: the matrix $A$ is diagonalizable if there's an invertible matrix $P$ such that $$A=PDP^{-1}$$ where $D=\mathrm{diag}(\lambda_1,\cdots,\lambda_n)$ is a diagonal matrix but this means that $P$ is the change of base matrix so there's a basis $\mathcal B=(v_1,\ldots,v_n)$ in which $A$ is diagonal: $Dv_i=\lambda_i v_i$ hence $v_i$ is an eigenvector of $D$ associated to the eigenvalue $\lambda_i$ so $\mathcal B$ is a basis formed by a linearly independent set of eigenvectors.

$\endgroup$
1
$\begingroup$

The matrix of a linear operator $T$ on a basis $B=(b_1,\ldots,b_n)$ is diagonal if and only if the basis vectors $b_1,\ldots,b_n$ are all eigenvectors of$~T$. This is immediate from the definitions eigenvector and a diagonal matrix (just write out what applying $T$ ot $b_i$ does).

A linear operator$~T$ on$~V$ is diagonalisable if on some basis of$~V$ its matrix is diagonal. By the above, that means precisely that there exists a basis of$~V$ consisting of eigenvectors of$~T$, and on such a basis the matrix of$~T$ is diagonal. So unlike what you said in the last paragraph of the question, having $\dim V$ linearly independent eigenvectors is a necessary and sufficient condition for $T$ to be diagonalisable

If the matrix$~A$ of$~T$ is given on some (other) basis, then the condition that $T$ (and hence$~A$) is diagonalisable is that some basis transformation (namely one from the given basis to a basis of eigenvectors) will bring $A$ to diagonal form. A basis transformation is $A\mapsto P^{-1}AP$ for some invertible matrix$~P$ which gives your "other" definition of diagonalisable. It has the drawback of hiding the geometric meaning somewhat, and not being applicable to linear operators without first expressing them on a basis by a matrix.

$\endgroup$
1
$\begingroup$

Let $\beta=\{v_1,v_2,\dots,v_n\},\beta_0$ is the standard basis.

From the definition of diagonality, if $A=Q^{-1}DQ$, $A$ is diagonalizable.

$[Av_i]_\beta=[\lambda_iv_i]_\beta=\lambda_ie_i$ (change the basis of eigenvectors $v_i$), so $[A]_\beta=[\lambda_1 e_1,\lambda_2 e_2,\dots,\lambda_i e_i]$, which is a diagonal matrix, as the question claimed.

Additionally, let $Q=[I]_{\beta_0}^{\beta}$, $[A]_{\beta_0}=Q^{-1}[A]_\beta Q$, and it is diagonalizable.

Notes: $A$ is diagonalizable $\iff A$ has $n$ linearly independent eigenvectors

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.