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Let $f_k \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear map defined by $$ f_k \begin{pmatrix} 1 \\ 2 \\ k \end{pmatrix}= \begin{pmatrix} 2+k \\ 3 \\ 0 \end{pmatrix}, \quad f_k \begin{pmatrix} 2 \\ k+1 \\ -1 \end{pmatrix}= \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix}, \quad f_k \begin{pmatrix} -3 \\ 1 \\ 5 \end{pmatrix}= \begin{pmatrix} 1 \\ k \\ 2 \end{pmatrix}, $$ How many functions $f_k$ are there?

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The solution are: infinite for $k=-4$, zero for $k = 2/3$ and exactly one in all other case. –  Katy23 May 29 '11 at 21:42
    
@Katy23: So, are you asking how to obtain those answers? And are they required to be linear transformations? –  Arturo Magidin May 29 '11 at 21:45
    
Yes. I don't know how to obtain those answers :(! –  Katy23 May 29 '11 at 21:47
    
@Katy23: Sorry, I just edited; are they also required to be linear transformations? –  Arturo Magidin May 29 '11 at 21:47
    
Yes :). I change the question text to add this information –  Katy23 May 29 '11 at 21:49

2 Answers 2

up vote 3 down vote accepted

First, consider the three vectors, $$\left(\begin{array}{c}1\\2\\k\end{array}\right),\qquad \left(\begin{array}{c}2\\k+1\\-1\end{array}\right),\qquad \left(\begin{array}{c}-3\\1\\5\end{array}\right).$$ If they are linearly independent, then we can define a linear transformation from $\mathbb{R}^3$ to any vector space simply by saying what we want the value to be at those three vectors (since any specification of what happens to a basis gives a linear transformation). If the three vectors are linearly dependent, then any linear dependency between them must also be satisfied by their images.

So, when are they linearly independent? The three vectors are linearly independent if and only if the determinant of the $3\times 3$ matrix that has them as columns is nonzero. Since $$\begin{align*} \det\left(\begin{array}{ccc}1 & 2 & -3\\ 2 & k+1 & 1\\ k & -1 & 5 \end{array}\right) &= \left|\begin{array}{cc}k+1 & 1\\ -1 & 5\end{array}\right| -2\left|\begin{array}{cc}2 & 1\\k & 5\end{array}\right| -3\left|\begin{array}{cc}2 & k+1\\k & -1 \end{array}\right|\\ &= 5(k+1) + 1 -2\Bigl(10 - k\Bigr) -3\Bigl(-2 -k^2-k\Bigr)\\ &= 5k + 5 +1 - 20 + 2k +6 +3k^2 + 3k\\ &= 3k^2 +10k -8 = (k+4)(3k-2). \end{align*}$$ So the three vectors are linearly independent if $(k+4)(3k-2)\neq 0$; that is, if $k\neq -4$ and $k\neq\frac{2}{3}$. In those cases, such a linear transformation exists, and since we've specified what happens at a basis, there is one and only one such linear transformation.

If $k=-4$, then the vectors are $$\left(\begin{array}{r}1\\2\\-4\end{array}\right),\qquad \left(\begin{array}{c}2\\-3\\-1\end{array}\right),\qquad \left(\begin{array}{c}-3\\1\\5\end{array}\right).$$ Notce that if we add all three vectors we get $\mathbf{0}$. That means that we must have that when we add the three images we get $\mathbf{0}$ as well, otherwise, we don't have a linear transformation; this holds, since the images are $$\left(\begin{array}{r}-2\\3\\0\end{array}\right),\qquad \left(\begin{array}{r}1\\1\\-2\end{array}\right),\qquad \left(\begin{array}{r}1\\-4\\2\end{array}\right).$$ In fact, this is (up to multiplying by a scalar) the only linear relation between the three vectors: they span a 2-dimensional subspace. Add a single vector to, say, the first two, to get a basis. Then you can define a linear transformation to be what we want at these three, and anything whatsoever on the third basis basis vector. Each choice will give you a linear transformation; so there are infinitely many possible linear transformations when $k=-4$.

Finally, when $k=\frac{2}{3}$, we have that the three vectors are $$\left(\begin{array}{r}1\\2\\\frac{2}{3}\end{array}\right),\qquad \left(\begin{array}{r}2\\\frac{5}{3}\\-1\end{array}\right),\qquad \left(\begin{array}{r}-3\\1\\5\end{array}\right).$$ The three are linearly dependent; if you do a bit of work, you will find that $$3\left(\begin{array}{c}1\\2\\ \frac{2}{3}\end{array}\right) - 3\left(\begin{array}{r}2\\ \frac{5}{3}\\-1\end{array}\right) -\left(\begin{array}{r}-3\\1\\5\end{array}\right) = \left(\begin{array}{c}3 -6 +3\\ 6-5-1\\2+3-5 \end{array}\right) = \left(\begin{array}{c}0\\0\\0\end{array}\right).$$ That means that we need the images to satisfy the same linear relation; but $$3\left(\begin{array}{c}2+\frac{2}{3}\\3\\0\end{array}\right) -3\left(\begin{array}{c}1\\1\\-2\end{array}\right) - \left(\begin{array}{c}1\\\frac{2}{3}\\2\end{array}\right)\neq\left(\begin{array}{c}0\\0\\0\end{array}\right),$$ so there can be no linear transformation with those values when $k=\frac{2}{3}$. a

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Great answer! Just a little clarification: I think the first matrix mentioned should have $1$ as its left, uppermost entry instead of $2+k$. –  Fernando Martin May 30 '11 at 1:39
    
@fmartin: Yes, thank you; messed up copying. –  Arturo Magidin May 30 '11 at 1:41

As to how to find them, just plug values of $k$ to the basis $\{(1,2,k), (2,k+1,-1),(-3,1,5)\}$. You can check that those three vectors are linearly independent for all $k\neq 4 \neq 2/3$, so those define a unique linear map since they span all the domain. For $k=4$, you can check that the map defined is linear (that is, $f(\alpha u+\beta v)) = \alpha f(u)+\beta f(v), \alpha, \beta \in \mathbb{Z}$), and so you can define an infinite amount of linear maps giving a third vector that completes a basis of $\mathbb{R}^3$; $k=2/3$ doesn't verify that $f(\alpha u+\beta v)=\alpha f(u)+\beta f(v)$ for some of the three vectors, so there's no way to give a linear map that verifies what's being asked.

In particular, for $k=2/3$: $3(1,2,2/3) - 3(2,5/3,-1) = (-3,1,5)$. If the map were linear, then $f(-3,1,5) = 3f(1,2,2/3) - 3f(2,5/3,-1) = (4,7,-6)$, but the question says it should be $(1,2/3,2)$

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i already found your results. my difficulty is how to set the problem. –  Katy23 May 29 '11 at 21:55

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