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.

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#.