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$$ A = \begin{pmatrix} λ & 1 & 1 \\ 1 & λ & 1 \\ 1 & 1 & λ \end{pmatrix} $$ My attempt:

For Rank(A) = 1, the rows of matrix A must be scalar multiples of each other. Thus, the only value that works in this case would be λ = 1.

For Rank(A) = 2, the determinant of A must equal zero in order to find the values of λ.

The determinant is λ^3 -3λ +2 = 0 which gives values of λ = -2 and λ = 1. Since we know that λ = 1 makes Rank(A) = 1, we can disregard λ = 1 in this case. Thus λ = -2 would make Rank(A) = 2.

For Rank(A) = 3, the determinant of A must not equal zero. Since we found values of λ for which det(A) = 0, the values of λ for which Rank(A) = 3 should be numbers that do not equal 1 or -2.

Am I missing any λ values for each case?

Andrew Nguyen
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4 Answers4

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Looks good to me. Note that this is a special case of a rank-1 update of a matrix.

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user1551
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If the rank is not three, then it is not invertible and hence, its determinant must be $0$ (iff). Finding the characteristi polynomial, we can see when this happens. $$\left\vert\begin{pmatrix} λ & 1 & 1 \\ 1 & λ & 1 \\ 1 & 1 & λ \end{pmatrix}\right\vert=\lambda\cdot\left\vert\begin{pmatrix} \lambda & 1 \\ 1 & \lambda\end{pmatrix}\right\vert-\left\vert\begin{pmatrix} 1 & 1\\1&\lambda\end{pmatrix}\right\vert+\left\vert\begin{pmatrix} 1 & \lambda\\1&1\end{pmatrix}\right\vert=\lambda(\lambda^2-1)-\lambda+1+1-\lambda=\lambda^3-\lambda+2-2\lambda=\lambda^3-3\lambda+2=(\lambda-1)^2(\lambda+2)$$ which implies $\lambda=1,-2$ as you stated. Now, check these two values if you need to to see which give rank to be $1$ and which give rank to be $2$. All other values clearly don't give determinant equal to $0$ and hence give rank=$3$. So, your reasoning was just about right. Good job.

user12345
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We could also try rowreduction. In order to avoid a casedistinction in the first step, I will switch row $1$ for row $3$. This gives us the matrix $$\begin{pmatrix} 1 & 1 & \lambda\\ 1 & \lambda & 1\\ \lambda & 1 & 1 \end{pmatrix}$$ and replacing row $2$ by (row $2$ - row $1$) and row $3$ by (row $3$ - $\lambda$ row $1$) gives us the following matrix: $$\begin{pmatrix} 1 & 1 & \lambda\\ 0 & \lambda - 1 & 1 - \lambda\\ 0 & 1 - \lambda & 1 - \lambda^2 \end{pmatrix}.$$ We now replace row $3$ by (row $3$ + row $2$) and find the matrix $$\begin{pmatrix} 1 & 1 & \lambda\\ 0 & \lambda - 1 & 1 - \lambda\\ 0 & 0 & -\lambda^2 - \lambda +2 \end{pmatrix}$$ and we have that $-\lambda^2 - \lambda + 2 = (1 - \lambda)(2 + \lambda)$. Therefore, we need to check the values $\lambda = 1$ and $\lambda = -2$. For $\lambda = 1$, we find by plugging this value in the last matrix: $$\begin{pmatrix} 1 & 1 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ so for $\lambda = 1$ the matrix has rank $1$. For $\lambda = -2$, we find the following matrix: $$\begin{pmatrix} 1 & 1 & -2\\ 0 & -3 & -1\\ 0 & 0 & 0 \end{pmatrix}$$ hence for $\lambda = -2$, the matrix has rank $2$. Rank $3$-matrices are obtained for any $\lambda \not\in \{-2, 1\}$.

$\textbf{REMARK:}$ Although this approach is clearly longer than yours, I find it more 'systematic' (personal opinion), although yours clearly uses some clever tricks. If you would ever use 'my' approach, switching rows etc can avoid case distinctions in early stages, making the work easier.

Student
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Your solution looks good.

Here’s another approach that also takes advantage of the special form of the matrix. It’s fairly easy to find without computing the characteristic polynomial of the matrix and finding its roots that the eigenvalues are $x+2$ with multiplicity one and $x-1$ with multiplicity two. (See here for an explanation.) The matrix is real symmetric, hence diagonalizable, so algebraic multiplicity equals geometric multiplicity for all of its eigenvalues. From this we can see immediately that the rank of the matrix is $1$ when $\lambda=1$, two when $\lambda=-2$ and three otherwise.

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amd
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