Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm given 4 vectors: $u_1, u_2, u_3$ and $u_4$. I'm going to type them in as points, because it will be easier to read, but think as them as column vectors.

$$u_1 =( 5, λ, λ, λ), \hspace{10pt} u_2 =( λ, 5, λ, λ), \hspace{10pt} u_3 =( λ, λ, 5, λ), \hspace{10pt}u_4 =( λ, λ, λ, 5)$$

The task is to calculate the value of λ if the vectors where linearly dependent, as well as linearly independent.

I managed to figure out that I could put them in a matrix, let's call it $A$, and set $det(A) = 0$ if the vectors should be linearly dependent, and $det(A) \neq 0$ if the vectors should be linearly independent.

Some help to put me in the right direction would be great!

share|cite|improve this question
Compute the determinant to get a polynomial in the variable $\lambda$; find out for what $\lambda$'s the polynomial is zero vs. nonzero (corresponding to linearly dependent and independent respectively). – anon Sep 26 '12 at 21:14
@anon: That's a lot more effort than is required. – joriki Sep 26 '12 at 21:28

If you write it as a matrix you will see the answer immediately...

For a certain value of lambda all the vectors will be equal and thus linearly dependent. What is it?

For another value of lambda the matrix will be a scalar times the identity matrix and thus linearly independent. What is this value?

share|cite|improve this answer
Well, if lambda were equal to 3 all the vectors would be equal. Is that the case? Thanks for helping out – marsrover Sep 29 '12 at 12:19

Here is a short cut. Form the $4\times 4$ matrix of all ones, call it $J$. Then your matrix can be represented as $$\lambda J - (\lambda - 5)I$$ $J$ is symmetric hence (orthogonally) diagonalizable. It's easy to see that the matrix has rank $1$ and one of the eigenvalues is $4$. Therefore the diagonal form is $\mathrm{diag}(4,\ 0,\ 0,\ 0)$. Thus we are reduced to calculating the determinant of the diagonal matrix $$\mathrm{diag}(3\lambda + 5,\ 5-\lambda,\ 5-\lambda,\ 5-\lambda)$$ It is then easy to see that the vectors will be linearly dependent if and only if $\lambda = 5$ or $\lambda = \frac{-5}{3}$

share|cite|improve this answer
nice observation and solution! – Tapu Sep 26 '12 at 21:39

If $\lambda=0$, the vectors are clearly linearly independent.

If $\lambda\ne0$, we can divide through by $\lambda$ without affecting whether the determinant vanishes; this yields

$$ \pmatrix{\frac5\lambda&1&1&1\\1&\frac5\lambda&1&1\\1&1&\frac5\lambda&1\\1&1&1&\frac5\lambda}\;. $$

Thus the values of $\lambda$ for which the determinant vanishes are those for which

$$ \frac5\lambda=1-\mu_i\;, $$

where $\mu_i$ is an eigenvalue of

$$ \pmatrix{1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1}\;. $$

This matrix annihilates all vectors whose components sum to $0$, so it has an eigenspace of dimension $4-1=3$ corresponding to the eigenvalue $0$, and thus a triple eigenvalue $0$. Since it is symmetric, its four eigenvectors can be chosen to form an orthonormal system, so the fourth eigenvector is a vector orthogonal to that eigenspace, e.g. $(1,1,1,1)$, which corresponds to eigenvalue $4$.

Thus we have the two possibilities $5/\lambda=1-0$, corresponding to $\lambda=5$, and $5/\lambda=1-4$, corresponding to $\lambda=-5/3$. All other values of $\lambda$ lead to linearly independent vectors.

share|cite|improve this answer

Your guess is correct. But what is the difficulty then?

May be you are unable to get the determinant in a simpler way..?

$$\left|\begin{array}{cccc} 5&\lambda&\lambda&\lambda\\\lambda&5&\lambda&\lambda\\\lambda&\lambda&5&\lambda\\\lambda&\lambda&\lambda&5\end{array}\right|=(5-\lambda)^3\left|\begin{array}{rrrr} 1&0&0&\lambda\\-1&1&0&\lambda\\0&-1&1&\lambda\\0&0&-1&5\end{array}\right|=(5-\lambda)^3\left[\left|\begin{array}{rrr}1&0&\lambda\\-1&1&\lambda\\0&-1&5 \end{array}\right|+\left|\begin{array}{rrr}0&0&\lambda\\-1&1&\lambda\\0&-1&5 \end{array}\right|\right]$$

So, finally, the determinant is $$(5-\lambda)^3\left[(5+\lambda)+\lambda+\lambda\right]=(5-\lambda)^3(5+3\lambda)$$

share|cite|improve this answer

Here is another way

$\left[\begin{matrix}5&\ell&\ell&\ell\\\ell & 5&\ell&\ell\\\ell&\ell&5&\ell\\\ell&\ell&\ell&5\end{matrix}\right]\sim\left[\begin{matrix}5 &\ell&\ell&\ell\\0&5-\ell &\ell-5&0\\0&0&5-\ell&\ell-5\\\ell&\ell&\ell&5\end{matrix}\right]\sim\left[\begin{matrix}5 &\ell&\ell&\ell\\0&5-\ell &\ell-5&0\\0&0&5-\ell&\ell-5\\\ell&\ell&\ell&5\end{matrix}\right]\sim\left[\begin{matrix}5 &\ell&\ell&\ell\\0&5-\ell &\ell-5&0\\0&0&5-\ell&\ell-5\\\ell-5&0&0&5-\ell\end{matrix}\right]\sim\left[\begin{matrix}5+\ell &\ell&\ell&\ell\\0&5-\ell &\ell-5&0\\\ell-5&0&5-\ell&\ell-5\\0&0&0&5-\ell\end{matrix}\right] $


$\left|\begin{matrix}5+\ell &\ell&\ell&\ell\\0&5-\ell &\ell-5&0\\\ell-5&0&5-\ell&\ell-5\\0&0&0&5-\ell\end{matrix}\right|=(5-\ell)\left|\begin{matrix}5+\ell&\ell&\ell\\0&5-\ell&\ell-5\\\ell-5&0&5-\ell\end{matrix}\right|=(5-\ell)\left|\begin{matrix}5+2\ell&\ell&\ell\\\ell-5&5-\ell&\ell-5\\0&0&5-\ell\end{matrix}\right|=(5-\ell)^{2}\left|\begin{matrix}5+2\ell&\ell\\\ell-5&5-\ell\end{matrix}\right|=(5-\ell)^3(5+3\ell)$

and see for which values of λ the determinant is zero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.