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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!

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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

5 Answers 5

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?

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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 $\lambda = \frac{-5}{3}$

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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.

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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 $$(5-\lambda)^3\left[(5+\lambda)+\lambda+\lambda\right]=(5-\lambda)^3(5+3\lambda)$$

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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] $

so

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

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

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