Compute the minimal polynomial of the matrix, without computing the characteristic polynomial.
$$
\begin{bmatrix}
4 & -4 & -4 & 0 \\
-5 & 7 & 4 & 0 \\
-4 & 0 & 3 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
$$
To compute the mipoly(X) of matrix M, we could suppose the mipoly has degree d, where 1<=d<=4. We have 4 cases when d = 1, d = 2, d=3, d=4
If d = 1, then mipoly = x + b, therefore, M + b*Id = 0 but this is not true.
If d = 2, then mipoly = xˆ2 + ax + b, therefore, Mˆ2 + aM + bId = 0 after doing some calculations, I came to the conclusion that the system has no non-zero solutions since the rank of the matrix is 3.
If d = 3, then Mˆ3 + aMˆ2 + bM + c*Id = 0 I computed the cube and the square and then formed one big matrix of the four matrices multiplied by 1,a,b,c, some real numbers to be equal to 0 vector.
$$
\begin{bmatrix}
604 & 52 & 4 & 1 \\
-516 & -44 & -4 & 0 \\
-516 & -44 & -4 & 0 \\
-516 & -44 & -4 & 0 \\
….
\end{bmatrix}
$$
multiplied by
$$
\begin{pmatrix}
1\\
a\\
b\\
c\\
\end{pmatrix}
$$
I formed a system of 64 linear equations and found c = 1, b = 4, a = 6 Could you please confirm with me whether you got the same values for a,b,c ?
Best Answer
So let me apply the technique of this answer to the current question. We first pick a nonzero vector whose repeated images by$~M$ are easy to compute; it is hard to resist taking the final standard basis vector$~e_4$ which is its own image. That gives that $Me_4-Ie_4=0$ and $P=X-1$ is the lowest degree monic polynomial such that $P[M]e_4=0$, so it divides the minimal polynomial. It is not so hard to see that $\lambda=1$ is not an eigenvalue of the top-left $3\times3$ submatrix, in other words that the image of $P[M]$ is the subspace spanned by $e_1,e_2,e_3$, and it remains to find the minimal polynomial of the restriction to that subspace (i.e., of that $3\times3$ submatrix).
In that subspace $e_2$ looks like a promising vector to compute repeated images of. $Me_2$ is nonzero and in the span of $e_1,e_2$, while $M^2e_2$ is not in the span of $e_1,e_2$, so no polynomial $Q$ of degree${}<3$ will have $Q[M]e_2=0$. There must be a monic such polynomial of degree$~3$, so lets write $M^3e_2=-516e_1+767e_2+224e_3$ as linear combination of $e_2$, $Me_2=-4e_1+7e_2$ and $M^2e_2=-44e_1+69e_2+16e_3$. The coefficient of $M^2e_2$ must be $224/16=14$, and subtracting $14M^2e_2$ from $M^3e_2$ leaves $100e_1-199e_2$ which is $-25Me_2-24e_2$, so $Q=X^3-14X^2+25X+24$. Indeed no root $1$ there, as expected, so your minimal polynomial is $PQ=X^4-15X^3+39X^2-X-24$.