## Practice Midterm SOLUTION: #2 |
abbyto1607 |
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Question two first asks for us to put the matrix A, $ \begin{array}{ccc} 1&2&3&3 \\ 2&2&4&7 \\ 4&1&0&1 \end{array} $ into row echelon form. In RREF, A is $ \begin{array}{ccc} 1&0&0&-7/6 \\ 0&1&0&17/3\\ 0&0&1&-1/2 \end{array} $. Now, we can use this to determine the dimension of the image of the linear transformation given by multiplication by A. We know that $rank(A)=dim(im(A))$. Since A in RREF has 3 leading ones, A is of rank 3, and $dim(im(A))=3$ To confirm this, we see that the last column in the RREF of matrix A corresponds to a redundant vector (since that column doesn't have a leading one). Thus, $im(A)$ is spanned by $ \begin{array}{ccc} 1 \\ 2 \\ 4 \end{array} $ $ \begin{array}{ccc} 1 \\ 2 \\ 1 \end{array} $ $ \begin{array}{ccc} 3 \\ 4\\ 0\end{array} $. Im(A) is spanned by three vectors and therefore dim(im(A))=3. From the rank-nullity theorem, we know that $dim(ker(A))+dim(im(A))=dim(A)$ From previous information and this theorem, $dim(ker(A))=4-3=1$ To find ker(A), we want to find all vectors that map to zero when multiplied by A. This is done by setting A to 0, row reducing, and solving for the variables. $ \begin{array}{ccc} 1&2&3&3&|0 \\ 2&2&4&7&|0 \\ 4&1&0&1&|0 \end{array} $ Row reducing, we get that $ \begin{array}{ccc} 1&0&0&-7/6 &|0\\ 0&1&0&17/3&|0\\ 0&0&1&-1/2 &|0\end{array} $. the vectors that map to zero are given by $ \begin{array}{ccc} -7/6 t\\ 17/3t\\ -1/2 t\end{array} $. where $t$ is arbitrary. Thus, ker(A) is spanned by $ \begin{array}{ccc} -7/6\\ 17/3\\ -1/2\end{array} $ |

swagata: May 5, 2015, 6:01 p.m. |
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I think the signs for the kernel should be the opposite of what is shown. For example, the top row of the matrix in RREF indicates that x - (7/6)w = 0. If w = t, then x = (7/6)t. So the elements in the span of the kernel should be 7/6, -17/3, and 1/2 (I think?). |

jlumxajaha: May 5, 2015, 8:34 p.m. |
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Isn't the top row of A supposed to be 1 1 3 3? Then wouldn't the final answer be ker(A) spanned by (I don't know how to make a vector with this website) -1/3, -25/6, 1/2, 1? In response to swagata, I don't think it matters if it's negative or positive because the opposite vector to any vector is exactly parallel to that vector so they're linearly dependent (or something). |

jlumxajaha: May 5, 2015, 8:40 p.m. |
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Oops I made an addition problem. I meant (1 -44/6 1/2 1) as the span of the kernel. Can someone check this? |

jlumxajaha: May 5, 2015, 8:46 p.m. |
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ROFL Okay I'm terrible at addition. But one of my other points still stands: shouldn't there be a 1 after -7/6 17/3 and -1/2? |

angela: May 5, 2015, 9:16 p.m. |
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I agree with @jlumxajaha. The kernel should have 4 rows. By definition, the kernel is all vectors that take the transformation to 0, so if you wrote it out, it would look like $\begin{bmatrix} 1 & 1 & 3 & 3 \\ 2 & 2 & 4 & 7 \\ 4 & 1 & 0 & 1 \end{bmatrix}$ $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ = $\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ with the kernel being whatever $\begin{bmatrix} a \\ b \\ c \\ d \end{bmatrix}$ is |

kcason: May 5, 2015, 10:15 p.m. |
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Agreed with @jlumxajaha/@angela (with the reasoning for the 4 rows) and @swagata (with the reasoning for switching the signs), except if it asks for all vectors that map to the 0 vector, shouldn't you include the arbitrary variable? As in the answer would be: 7/6t -17/3t 1/2t t where t is an element of all real numbers. |

abbyto1607: May 5, 2015, 10:31 p.m. |
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Yep, it should have four rows, with t as the last. Sorry for the mistake, I did the same thing with another problem. It was an early error I fixed, but forgot to when typing out the solution. |

leopham: May 5, 2015, 10:40 p.m. |
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I agree with swagata, the span of the kernel should be x_1_= 7/6, x_2_= -17/3, x_3_=1/2, and x_4_ = t. The vectors that map A to 0 would be kernel as Ker(t) represents the dimension that is lost during the linear transformation. |