Previously we saw that a number #\lambda# is an eigenvalue of #L# if and only if #\det (L-\lambda \cdot I_V)=0#. This statement can be refined by means of the notion of *minimal polynomial*.

Let #L:V\to V# be a linear map from a finite-dimensional vector space #V# to itself. A number #\lambda# is an eigenvalue of #L# if and only if it is a root of the *minimal polynomial* of #L#.

If #V# is a real vector space, then, according to the *Fundamental theorem of algebra*, the characteristic polynomial of #L# will factor into a product of linear polynomials and quadratic polynomials with leading coefficients #1# and negative discriminants. Each of these quadratic polynomials will also occur as a divisor of the minimal polynomial (not necessarily with the same multiplicity).

If #\lambda# is a root of the minimal polynomial, then it also is a root of the characteristic polynomial (which, after all, is a multiple of the minimal polynomial), and so, according to *rule 1 for the Characterization of eigenvalues and eigenvectors*, an eigenvalue.

Conversely, if #\lambda# is an eigenvalue of #L# with eigenvector #\vec{v}#, then, for the minimal polynomial #m_L(x)#, we have \[\begin{array}{rcl}m_L(\lambda)\,\vec{v}&=&m_L(L)\,\vec{v} \\&&\phantom{xx}\color{blue}{\lambda\,\vec{v} = L(\vec{v})}\\ &=&\vec{0}\\&&\phantom{xx}\color{blue}{m_L(L)=0_V}\end{array}\] Since #\vec{v}# is an eigenvector, we have #\vec{v}\ne\vec{0}#, so #m_L(\lambda) = 0#. This shows that #\lambda# is a root of the minimal polynomial.

It remains to prove the last statement. Suppose that #V# is a real vector space and that #p(x)# is a quadratic polynomial with leading coefficient #1# and negative discriminant which divides the characteristic polynomial of #L#. If #\lambda# is a root of #p(x)#, then it is not real, and therefore the complex conjugate #\overline{\lambda}# is the second root of #p(x)#. By applying the above to the *complexification* of #V#, we see that both #x-\lambda# and #x-\overline{\lambda}# occur as factors of the minimal polynomial of #L# (the minimal polynomial of #L# on the complexification of #V# equals the minimal polynomial of #L# on #V#). Because #\lambda\ne\overline{\lambda}#, the product #p(x) = (x-\lambda) \cdot (x-\overline{\lambda})# is a divisor of the minimal polynomial.

The difference between the minimal polynomial and the characteristic polynomial is that factors occurring more than once in the characteristic polynomial may occur with a lower multiplicity in the minimal polynomial. The following three matrices illustrate this.

\[ J_{1} = \matrix{0&0&0\\ 0&0&0\\ 0&0&0},\quad J_{2} = \matrix{0&1&0\\ 0&0&0\\ 0&0&0},\quad J_{3} = \matrix{0&1&0\\ 0&0&1\\ 0&0&0}\]

All three have characteristic polynomial #x^3#. But

- #J_{1}# has minimal polynomial #x#,
- #J_{2}# has minimal polynomial #x^2#,
- #J_{3}# has minimal polynomial #x^3#.

These matrices are examples of matrices in *Jordan normal form*, which we will discuss later.

We use the minimal polynomial for the following characterization of diagonalizability.

Let #V# be a vector space of finite dimension #n# with basis #\alpha# and let #L:V\to V# be a linear map.

- If #V# is a real vector space, then #L# is
*diagonalizable* (over the real numbers) if and only if the minimal polynomial of #L# is a product of a constant and linear factors (with leading coefficients equal to #1#), which are all mutually different.
- If #V# is a real vector space, then #L# is diagonalizable over the complex numbers if and only if every factor of the minimal polynomial of #L# (written with leading coefficient equal to #1#) occurs only once.
- If #V# is a complex vector space, then #L# is diagonalizable if and only if the minimal polynomial of #L# has no double roots.

If #L_\alpha# is a diagonal matrix, and #\lambda_1,\ldots,\lambda_m# are the mutually different numbers on its diagonal, then the minimal polynomial of #L_\alpha#, and thus of #L#, is equal to the product \[(x-\lambda_1)\cdots(x-\lambda_m)\]

Conversely, if, for mutually different numbers #\lambda_1,\ldots,\lambda_m#, the minimal polynomial of #L# is equal to #m_L(x) = (x-\lambda_1)\cdots(x-\lambda_m)#, then, because of the above theorem* Roots of the minimal polynomial*, these numbers are eigenvalues of #L#. Furthermore, with #c_i =\prod_{j\ne i} \frac{1}{\lambda_i-\lambda_j}#, we have

\[1= \sum_{i=1}^m c_i \prod_{j\ne i}(x-\lambda_j)\]

This follows from the fact that the right-hand side is a polynomial of degree #m-1# with the value #1# at #m# distinct points #x=\lambda_1,\ldots,\lambda_m# (see *Lagrange's theorem).*

We deduce that #V# is the *direct sum* of the *eigenspaces* \( E_i = \im{\prod_{j\ne i}(L-\lambda_j\cdot I_V)}\).

To this end, we first substitute #L# into the formula above, and study the image of an arbitrary vector #\vec{v}# under the result:

\[\begin{array}{rcl} \vec{v} &=& I_V(\vec{v})\\ &=&\displaystyle\sum_{i=1}^m c_i \prod_{j\ne i}(L-\lambda_j\cdot I_V)(\vec{v})\\ &\in& E_1+\cdots+E_m\end{array}\] This shows that #V# is the sum of the linear subspaces #E_i#.

We also see that

\[\begin{array}{rcl}(L-\lambda_i\cdot I_V )(E_i) &=&\displaystyle(L-\lambda_i\cdot I_V )\left( \im{\prod_{j\ne i}(L-\lambda_j\cdot I_V)}\right) \\&=&\displaystyle\im{\prod_{j}(L-\lambda_j\cdot I_V)}\\& = &\im{m_L(L)}\\& = &\im{0_V}\\ & =& \{\vec{0}\}\end{array}\] so #E_i# is contained in #\ker{L-\lambda_i\cdot I_V}#.

If, for each #i#, a basis of #E_i# is chosen, then, according to theorem *Independence of eigenvectors for different eigenvalues*, the union of these bases is a set of linearly independent vectors. By what we saw above, these vectors span #V#. Therefore, this union is a basis for #V#. This shows that #V# is the direct sum of the eigenspaces #E_i#.

This proves the first and the last statement. The second statement follows by applying this result to the complexification of #V#.

The matrix \[A = \matrix{1&1\\ 0&1}\] is not diagonalizable. For, otherwise there would be numbers #a# and #b# such that #A# is conjugate to #D=\matrix{a&0\\ 0&b}#. But then

\[\begin{array}{rclclclcl}a+b &=&\text{tr}(D) &=& \text{tr}(A )&=&1+1 &=& 2 \\ a\cdot b &=&\det( D) &=& \det(A) &=&1\cdot 1 -1\cdot 0 &=& 1 \end{array}\] If we substitute #b = 2 - a# from the first equation into the second equation, we find the quadratic equation #a^2-2 a+1=0#, which has only one solution: #a =1#. This implies #b = 1#, so #D = I_2#, the identity matrix. That means that there is an invertible #(2\times2)#-matrix #T# with #A = T \,I_2T^{-1}=I_2 #. This contradicts the fact that the #(1,2)#-entry of #A# is equal to #1#.

In accordance with the theorem, the minimal polynomial #(x-1)^2# of #A# has a double root.

- An orthogonal projection #P# onto a subspace of an inner product space #V# satisfies the equation #P^2=P# and so is diagonalizable.
- A reflection #S# about a linear subspace of dimension #n-1# in an #n#-dimensional inner product space #V# satisfies the equation #S^2 = I_V#, and so is diagonalizable.

A well-known criterion for the absence of multiple factors in the factorization of a polynomial #f(x)# is \[ \gcd(f(x),f'(x)) = 1\] Here #\gcd# stands for "greatest common divisor". With the aid of the *Euclidean algorithm for polynomials* this greatest common divisor can be found in an efficient manner. This leads to the following method for determining the diagonalizability of a square matrix #A#:

- Determine the minimal polynomial #m_A(x)# of #A#.
- Calculate #\gcd(m_A(x),\frac{\dd}{\dd x}(m_A(x))) #.
- If this #\gcd# equals #1#, then #A# is diagonalizable over the complex numbers. If so and if #A# is real, then #A# is diagonalizable over the real numbers if and only if all of the roots of #m_A(x)# are real.

Let #V# be a real vector space of dimension #6# and suppose that #L:V\to V# is a linear map with minimal polynomial

\[m_L(x ) =x^3+x^2-14 x-24 \] Is #L# diagonalizable over the real numbers?

Solution Yes

To see this, we determine the eigenvalues of #L#. Trial and error yields that #-3# is a root of \(m_L(x )\). Division by #x+3# gives the factorization

\[m_L(x ) = (x+3) \cdot ( x^2-2 x-8)\]The quadratic factor factors as #(x-4)\cdot (x+2)#, so the minimal polynomial has three distinct real roots. According to the first item of theorem

*Recognizing diagonalizability using the minimal polynomial* the answer is therefore: Yes.