Diagonalizable matrix

In linear algebra, a square matrix $A$ is called diagonalizable or non-defective if it is similar to a diagonal matrix. That is, if there exists an invertible matrix $P$ and a diagonal matrix $D$ such that . This is equivalent to (Such $D$ are not unique.) This property exists for any linear map: for a finite-dimensional vector space a linear map $T:V\backslash to\; V$ is called diagonalizable if there exists an ordered basis of $V$ consisting of of $T$. These definitions are equivalent: if $T$ has a matrix representation $A\; =\; PDP^\{-1\}$ as above, then the column vectors of $P$ form a basis consisting of eigenvectors of and the diagonal entries of $D$ are the corresponding of with respect to this eigenvector basis, $T$ is represented by

Diagonalization is the process of finding the above $P$ and and makes many subsequent computations easier. One can raise a diagonal matrix $D$ to a power by simply raising the diagonal entries to that power. The determinant of a diagonal matrix is simply the product of all diagonal entries. Such computations generalize easily to

The geometric transformation represented by a diagonalizable matrix is an inhomogeneous dilation (or anisotropic scaling). That is, it can scale the space by a different amount in different directions. The direction of each eigenvector is scaled by a factor given by the corresponding eigenvalue.

A square matrix that is not diagonalizable is called Defective matrix. It can happen that a matrix $A$ with real number entries is defective over the real numbers, meaning that $A\; =\; PDP^\{-1\}$ is impossible for any invertible $P$ and diagonal $D$ with real entries, but it is possible with complex number entries, so that $A$ is diagonalizable over the complex numbers. For example, this is the case for a generic rotation matrix.

Many results for diagonalizable matrices hold only over an algebraically closed field (such as the complex numbers). In this case, diagonalizable matrices are Dense set in the space of all matrices, which means any defective matrix can be deformed into a diagonalizable matrix by a small perturbation; and the Jordan–Chevalley decomposition states that any matrix is uniquely the sum of a diagonalizable matrix and a nilpotent matrix. Over an algebraically closed field, diagonalizable matrices are equivalent to semi-simple matrices.

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Definition A square $n\; \backslash times\; n$ matrix $A$ with entries in a field $F$ is called diagonalizable or nondefective if there exists an $n\; \backslash times\; n$ invertible matrix (i.e. an element of the general linear group GL _n( F)), $P$, such that $P^\{-1\}AP$ is a diagonal matrix.

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Characterization The fundamental fact about diagonalizable maps and matrices is expressed by the following:

• An $n\; \backslash times\; n$ matrix $A$ over a field $F$ is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to $n$, which is the case if and only if there exists a basis of $F^n$ consisting of eigenvectors of $A$. If such a basis has been found, one can form the matrix $P$ having these basis vectors as columns, and $P^\{-1\}AP$ will be a diagonal matrix whose diagonal entries are the eigenvalues of $A$. The matrix $P$ is known as a modal matrix for $A$.
• A linear map $T\; :\; V\; \backslash to\; V$ is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to which is the case if and only if there exists a basis of $V$ consisting of eigenvectors of $T$. With respect to such a basis, $T$ will be represented by a diagonal matrix. The diagonal entries of this matrix are the eigenvalues of

The following sufficient (but not necessary) condition is often useful.

• An $n\; \backslash times\; n$ matrix $A$ is diagonalizable over the field $F$ if it has $n$ distinct eigenvalues in i.e. if its characteristic polynomial has $n$ distinct roots in however, the converse may be false. Consider $$\backslash begin\{bmatrix\}$$

-1 & 3 & -1 \\ -3 & 5 & -1 \\ -3 & 3 & 1 \end{bmatrix}, which has eigenvalues 1, 2, 2 (not all distinct) and is diagonalizable with diagonal form (similar to and change of basis matrix $P$: The converse fails when $A$ has an eigenspace of dimension higher than 1. In this example, the eigenspace of $A$ associated with the eigenvalue 2 has dimension 2.

• A linear map $T\; :\; V\; \backslash to\; V$ with $n\; =\; \backslash dim(V)$ is diagonalizable if it has $n$ distinct eigenvalues, i.e. if its characteristic polynomial has $n$ distinct roots in $F$.

Let $A$ be a matrix over If $A$ is diagonalizable, then so is any power of it. Conversely, if $A$ is invertible, $F$ is algebraically closed, and $A^n$ is diagonalizable for some $n$ that is not an integer multiple of the characteristic of then $A$ is diagonalizable. Proof: If $A^n$ is diagonalizable, then $A$ is annihilated by some polynomial which has no multiple root (since and is divided by the minimal polynomial of

Over the complex numbers $\backslash Complex$, almost every matrix is diagonalizable. More precisely: the set of complex $n\; \backslash times\; n$ matrices that are not diagonalizable over considered as a subset of has Lebesgue measure zero. One can also say that the diagonalizable matrices form a dense subset with respect to the Zariski topology: the non-diagonalizable matrices lie inside the vanishing set of the discriminant of the characteristic polynomial, which is a hypersurface. From that follows also density in the usual ( strong) topology given by a norm. The same is not true over

The Jordan–Chevalley decomposition expresses an operator as the sum of its semisimple (i.e., diagonalizable) part and its nilpotent part. Hence, a matrix is diagonalizable if and only if its nilpotent part is zero. Put in another way, a matrix is diagonalizable if each block in its Jordan form has no nilpotent part; i.e., each "block" is a one-by-one matrix.

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Diagonalization Consider the two following arbitrary bases $E\; =\; \backslash$

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