求助：求解4×4矩阵A的特征值，计算过程遇阻

Hey there, I totally get how frustrating it is to spend an hour and a half grinding through a 4x4 eigenvalue problem without landing on clean results. Let's walk through a simpler approach to find the eigenvalues of your matrix:

$$
A = \begin{bmatrix} 7 & 1 & 2& 2\ 1&4&-1&-1\ -2&1&5&-1\ 1 &1&2 &8 \end{bmatrix}
$$

Instead of pushing forward with the transformed matrix you have, let's use row operations to simplify the characteristic matrix $A-\lambda I$ first—this will cut down on the messy algebra drastically.

First, write out $A-\lambda I$:
$$
A-\lambda I = \begin{bmatrix} 7-\lambda & 1 & 2& 2\ 1&4-\lambda&-1&-1\ -2&1&5-\lambda&-1\ 1 &1&2 &8-\lambda \end{bmatrix}
$$

Apply these row operations (they don't alter the determinant value):

• Row 1 = Row 1 - $(7-\lambda) \times$ Row 4
• Row 2 = Row 2 - $1 \times$ Row 4
• Row 3 = Row 3 + $2 \times$ Row 4

After these steps, we get:
$$
\begin{bmatrix} 0 & \lambda-6 & 2(\lambda-6) & -(\lambda-6)(\lambda-9)\ 0 & 3-\lambda & -3 & \lambda-9\ 0 & 3 & 9-\lambda & 15-2\lambda\ 1 & 1 & 2 & 8-\lambda \end{bmatrix}
$$

Now expand the determinant along the first column (only the bottom element is non-zero, which makes this trivial):
The determinant equals $1 \times (-1)^{4+1} \times \det(M_{41})$, where $M_{41}$ is the 3x3 matrix formed by removing row 4 and column 1:
$$
M_{41} = \begin{bmatrix} \lambda-6 & 2(\lambda-6) & -(\lambda-6)(\lambda-9)\ 3-\lambda & -3 & \lambda-9\ 3 & 9-\lambda & 15-2\lambda \end{bmatrix}
$$

Notice every element in the first row shares a common factor of $(\lambda-6)$—factor that out to simplify:
$$
\det(M_{41}) = (\lambda-6) \times \det\left(
\begin{bmatrix}
1 & 2 & -(\lambda-9)\
3-\lambda & -3 & \lambda-9\
3 & 9-\lambda & 15-2\lambda
\end{bmatrix}
\right)
$$

Next, simplify the 3x3 determinant by adding Row 1 to Row 2 (this eliminates the $\lambda-9$ term in Row 2, Column 3):
Row 2 = Row 2 + Row 1
$$
\begin{bmatrix}
1 & 2 & -(\lambda-9)\
4-\lambda & -1 & 0\
3 & 9-\lambda & 15-2\lambda
\end{bmatrix}
$$

Now expand this 3x3 determinant along the third column (the zero entry here saves us a lot of work):
$$
\begin{align*}
\det(\text{3x3 matrix}) &= -\left(\lambda-9\right) \times (-1)^{1+3} \times \det\left(\begin{bmatrix}4-\lambda & -1\3 &9-\lambda\end{bmatrix}\right) + \left(15-2\lambda\right) \times (-1)^{3+3} \times \det\left(\begin{bmatrix}1 &2\4-\lambda &-1\end{bmatrix}\right)
\end{align*}
$$

Calculate the two 2x2 determinants:

1. $\det\left(\begin{bmatrix}4-\lambda & -1\3 &9-\lambda\end{bmatrix}\right) = (4-\lambda)(9-\lambda) + 3 = \lambda^2 -13\lambda +39$
2. $\det\left(\begin{bmatrix}1 &2\4-\lambda &-1\end{bmatrix}\right) = -1 -2(4-\lambda) = 2\lambda -9$

Substitute these back in and simplify the expression:
$$
\begin{align*}
\det(\text{3x3 matrix}) &= -(\lambda-9)(\lambda^2 -13\lambda +39) + (15-2\lambda)(2\lambda -9)\
&= -\left(\lambda^3 -22\lambda^2 +156\lambda -351\right) + \left(-4\lambda^2 +48\lambda -135\right)\
&= -\lambda^3 +18\lambda^2 -108\lambda +216\
&= -\left(\lambda-6\right)^3
\end{align*}
$$

Putting all the pieces back together:

• $\det(M_{41}) = (\lambda-6) \times -\left(\lambda-6\right)^3 = -\left(\lambda-6\right)^4$
• $\det(A-\lambda I) = - \det(M_{41}) = \left(\lambda-6\right)^4$

So the characteristic polynomial is $\left(\lambda-6\right)^4$, which means the only eigenvalue of matrix $A$ is 6, with algebraic multiplicity 4. For extra context, the geometric multiplicity is 2 (since the rank of $A-6I$ is 2, leaving $4-2=2$ linearly independent eigenvectors).

That should get you past the messy algebra you were stuck on!

内容的提问来源于stack exchange，提问作者user522841