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I asked the same question yesterday, but this one is a bit different in terms of computations. It is from my exam I took an hour ago.

For what $h$ the columns of this matrix are linearly dependent? $$\begin{bmatrix} 1 & -3 & 4 \\ -4 & 7 & h\\ 2 & -6 & 8 \end{bmatrix}$$

Attempt: after row reducing, but not completely: $$\begin{bmatrix} 1 & -3 & 4 & 0 \\ -4 & 7 & h & 0 \\ 2 & -6 & 8 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 4 & 0 \\ 0 & -5 & h+16 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} $$

My guess was that if $h=-16;-\frac{28}{3}$ the system is linearly dependent. And I just guessed the -16. Hints please.

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Hint: the last row is all zeroes, independently of the value of $h$. – D. Thomine Jul 19 '12 at 20:00
Yeah so $x_3$ can be anything. But the vector that contains h should be collinear to at least one other vector. So, h should be some number(s) that lets its vector be a scalar multiple of another vector. Thats how i see it. – Koba Jul 19 '12 at 20:03
In fact, this is the same as your previous question (in spirit, if not in details): since the third row is a multiple of the first row, it doesn't matter what $h$ is, the matrix is not full rank. By the way: you did not give a set, you gave a matrix; and a matrix is neither linearly dependent nor independent. Presumably, you are refering to the rows or columns of the matrix in some way? – Arturo Magidin Jul 19 '12 at 20:07
In a square matrix, the rows are linearly independent if and only if the rows are linearly independent. This is not obvious, but it is nonetheless true: it follows from the fact that in any matrix, the dimension of the rowspace is the same as the dimension of the columnspace. But if you are interested in the linear independence of the columns, why are you looking at the extended matrix? It's not relevant. – Arturo Magidin Jul 19 '12 at 20:29
@Dostre: And why -16? Because that makes the entry equal to $0$? So what? That value comes from selecting $x_2=0$; but pick your favorite value of $x_2$, and then you can find a value of $h$ that will make the equation true. Conversely, pick your favorite value for $h$, solve for $x_2$ from $-5x_2 + h + 16 = 0$; and then use the value to solve for $x_1$ from $x_1-3x_2+4=0$ (having selected $x_3=1$): every value of $h$ gives you a solution. – Arturo Magidin Jul 19 '12 at 20:52
up vote 2 down vote accepted

You are asking: for what values of $h$ are the vectors $$\vec{v_1}=\left(\begin{array}{r}1\\-4\\2\end{array}\right),\quad \vec{v_2}=\left(\begin{array}{r}-3\\7\\-6\end{array}\right),\quad \vec{v_3}=\left(\begin{array}{r}4\\h\\8\end{array}\right)$$ linearly dependent?

You seem to be trying to do this by looking at the equation $$\alpha\vec{v_1}+\beta\vec{v_2}+\gamma\vec{v_3}=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$ and trying to determine for what values of $h$ there is a nonzero solution. This leads to the matrix you have: $$\left(\begin{array}{rrr|c} 1 & -3 & 4 & 0\\ -4 & 7 & h & 0\\ 2 & -6 & 8 & 0 \end{array}\right).$$ Now, since the third equation is a multiple of the first, that equation does not matter: it provides no new information. That means that you have a homogeneous system of two equations in three unknowns. Those systems always have infinitely many solutions. In particular, no matter what $h$ is, the system has infinitely many solutions, and so must have a nontrivial solution. Thus, the vectors are always linearly dependent.

To understand what is happening, note that all three vectors lie in the plane $z=2x$. Any two vectors on the plane that are not collinear will span the plane. Since $\vec{v_1}$ and $\vec{v_2}$ are not collinear, and both lie on the plane $z=2x$, any vector that lies on the plane $z=2x$ will be a linear combination of $\vec{v_1}$ and $\vec{v_2}$. Or, put another way, three vectors in a $2$-dimensional space (a plane through the origin) are always linearly dependent.

Here you have three vectors that satisfies $z=2x$; every other vector that satisfies that is a linear combination of $\vec{v_1}$ and $\vec{v_2}$: if $(a,b,2a)^t$ lies in the plane, then the system $$\alpha\left(\begin{array}{r}1\\-4\\2\end{array}\right) + \beta\left(\begin{array}{r}-3\\7\\-6\end{array}\right) = \left(\begin{array}{c}a\\b\\2a\end{array}\right)$$ has a solution, namely $\alpha = -\frac{7a+3b}{5}$, $\beta=-\frac{4a+b}{5}$ (obtained by Gaussian elimination). In particular, since no matter what $h$ is $\vec{v_3}$ lies in the plane $2z=x$, then we will have $$\vec{v_3} = -\frac{28+3h}{5}\vec{v_1} - \frac{16+h}{5}\vec{v_2}.$$ Note that this makes sense no matter what $h$ is.

This can be read off your row-reduced matrix: you got $$\left(\begin{array}{rrr|c} 1 & -3 & 4 & 0\\ 0 & -5 & h+16 & 0\\ 0 & 0 & 0 & 0 \end{array}\right).$$ Divide the second row by $-5$ to get $$\left(\begin{array}{rrr|c} 1 & -3 & 4 & 0\\ 0 & 1 & -\frac{h+16}{5} & 0\\ 0 & 0 & 0 & 0 \end{array}\right),$$ and now add three times the second row to the first row to get $$\left(\begin{array}{rrc|c} 1 & 0 & 4+\frac{-3h-48}{5} & 0\\ 0 & 1 & -\frac{h+16}{5} & 0\\ 0 & 0 & 0 & 0 \end{array}\right) = \left(\begin{array}{rrc|c} 1 & 0 & -\frac{28+3h}{5} & 0\\ 0 & 1 & -\frac{h+16}{5} & 0\\ 0 & 0 & 0& 0 \end{array}\right).$$ So $\alpha$ and $\beta$ are leading variables, and $\gamma$ is a free variable. This tells you that the solutions to the original system are: $$\begin{align*} \alpha &= \frac{28+3h}{5}t\\ \beta &= \frac{h+16}{5}t\\ \gamma &= t \end{align*}$$ Any nonzero value of $t$ gives you a nontrivial solution, and $t=-1$ gives you the solution I give above.

Of course, this can be done much more simply noting that since your original matrix has linearly dependent rows (third row is a scalar multiple of the first row), then the dimension of the rowspace is at most $2$ (in fact, exactly $2$), and hence the dimension of the columnspace is at most $2$ (in fact, exactly $2$, since $\dim(\text{columnspace})=\dim(\text{rowspace})$, so the columns are always linearly dependent.

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Based on your question from before, let's take a step back and revisit what linear dependence means.

Specifically, a set of vectors is linearly dependent if it is not linearly independent. Well maybe that's not useful, but now we get to talk about linear independence!

A set of vectors is linearly independent if none of the vectors in the set can be written as a linear combination of finitely many other vectors in the set. So, if you can write one of the vectors, say, $v_2$ as a combination of some finite number of other vectors, then your set is linearly dependent, e.g. $v_2 = 2v_1 + 5v_6$.

Rather than trying to compute things through row-reduction, which I suspect is confusing you, use the definition of linear independence to find a value of $h$ that makes the set not linearly independent.

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Compute the $det$, the root of $det=0$ is the only $h$ that makes the set linearly dependent.

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The first and third rows are the same, so the determinant will be zero for all $h$. – Emily Jul 19 '12 at 20:15
Then for all $h$, the columns are linearly dependent, since column rank = row rank – chaohuang Jul 19 '12 at 20:34

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