1

Let

$$A = \left[\begin{array}{rrrrr}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{array}\right] \in {M}_{5}(\mathbb{R})$$

Which of following options is $\det(A)$ ?

  1. $4^4 \times 14$

  2. $4^3 \times 14$

  3. $4^2 \times 14$

  4. $4 \times 14$

I think we have

$$\det \left[\begin{array}{ll}6 & 2 \\ 2 & 6\end{array}\right] = 4 \times 8$$

$$\det \left[\begin{array}{lll}6 & 2 & 2 \\ 2 & 6 & 2 \\ 2 & 2 & 6\end{array}\right] = 4^{2} \times10$$

and for any $n$ we have $\det(A_n)= 4^{n-1} \times (6+2×(n-1))$ so "1" is true.

Rodrigo de Azevedo
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amir bahadory
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  • Could you elaborate on what technique you think you are applying? Because I can't figure out what you're doing. – rschwieb Aug 07 '20 at 15:12
  • Consider $A - 4I$ and use https://en.wikipedia.org/wiki/Rank%E2%80%93nullity_theorem – Dmitry Aug 07 '20 at 15:13
  • For a proof, consider the eigenvectors and eigenvalues of $a\cdot \text{Ones}$ and $bI$ where $\text{Ones}$ is the matrix whose entries are all $1$ and how this relates to the egivenvectors and eigenvalues of $a\cdot \text{Ones}+bI$, the trace, and the determinant. – JMoravitz Aug 07 '20 at 15:13
  • (1) Add every other row to the first row; (2) Divide the first row by $6 + 2 \times (n - 1)$; (3) Subtract $2\times$ the first row from every other row. – WhatsUp Aug 07 '20 at 15:25
  • @rschwieb .i want to compute $det(A)$ for case $n=2,3 $ and find relation for every $n$ – amir bahadory Aug 07 '20 at 15:48

4 Answers4

4

As, dimension of nullspace of $(A-4I)$ is $4$. So, Geometric multiplicity of eigenvalue $4$ is $4$, as matrix $A$ is symmetric, so, $A$ must be diagonalizable, and hence, Algebraic multiplicity and Geometric multiplicity of eigenvalue $4$ is same, so, Algebraic multiplicity of eigenvalue $4$ is $4$.

And, as each row sum is $14$, so, $14$ is another eigenvalue of algebraic multiplicity $1$.

Det($A$)=multiplication of eigenvalues= $4^4×14$

A learner
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2

$$\begin{vmatrix}6 & 2 & 2 & 2 & 2 \\ 2 & 6 & 2 & 2 & 2 \\ 2 & 2 & 6 & 2 & 2 \\2 & 2 & 2 & 6 & 2 \\ 2 & 2 & 2 & 2 & 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 2 & 2 & 2 & 2 &\ \ 6\end{vmatrix}=\begin{vmatrix}4 & 0 & 0 & 0 & -4 \\ 0 & 4 & 0 & 0 & -4 \\ 0 & 0 & 4 & 0 & -4 \\0 & 0 & 0 & 4 & -4 \\ 0 & 0 & 0 & 0 & 14\end{vmatrix}$$

2

According to the Matrix Determinant Lemma, $\det(A+uv^T)= \det(A)(1+v^T A^{-1}u)$, hence

$$\det(A) = \det(4I + 2\cdot\mathbf 1 \mathbf 1^T) = \det(4I)(1+2\cdot \mathbf 1^T(\tfrac{1}{4}I)\mathbf 1) = 4^n (1 +\tfrac{1}{2}n) $$

For $n=5$, that's $4^5\cdot \frac{7}{2} = 4^4 \cdot 14$

Hyperplane
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1

Your generalization, with an obvious but unstated definition of $A_n$, can be proved as follows. Define $E_n$ as the $n\times n$ matrix whose entry are all $1$s. You can show its only nonzero eigenvalue is $2n$, and the associated eigenspace is $1$-dimensional. By the rank-nullity theorem, the eigenvalue $0$ has multiplicity $n-1$. So $A_n=4I_n+2E_n$ has eigenvalues $4,\,2n+4$ with respective multiplicities $n-1,\,1$.

J.G.
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