Linear Algebra Assignment #6

Solutions

1. Find a basis for the null space of the mapping $A\bar{x}$ when
   $$A=\left(\begin{array}{rrrrrr} 1&0&-2&0&1&0\\ 0&-2&3&0&2&-1\\ 0&0&0&1&-1&-2\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ \end{array}\right)$$
   What is the dimension of this subspace? Given this, determine the dimension of the subspace $Col\;A$. What is the rank of $A$?
   
   Solution:Write the solution set to $A\bar{x}=\bar{0}$ in parametric vector form. The set of all solutions will be of the form
   

   $$\bar{x}=\left(\begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ x_6\\ \end{array}\right) = \left(\begin{array}{c} 2x_3-x_5\\ 3x_3/2+x_5-x_6/2\\ x_3\\ x_5+2x_6\\ x_5\\ x_6\\ \end{array}\right)$$

   $$= x_3\left(\begin{array}{c} 2\\ 3/2\\ 1\\ 0\\ 0\\ 0\\ \end{array}\... ...\right)+x_6\left(\begin{array}{c} 0\\ 1/2\\ 0\\ 2\\ 0\\ 1\\ \end{array}\right)$$

   
   
   Since the null space is spanned by 3 linearly independent vectors it has dimension 3. Then since $dim(null\;A)+dim(Col\;A)=6$ the dimension of the column space, and therefore the rank, is three.
   

2. Find a basis for $Col\;A$ when
   $$A=\left(\begin{array}{rrrrr} 1&5&4&3&2\\ 1&6&6&6&6\\ 1&7&8&10... ...r} 1&0&-6&0&6\\ 0&1&2&0&-2\\ 0&0&0&1&2\\ 0&0&0&0&0\\ \end{array}\right)$$
   What is the dimension of this subspace? Given this, determine the dimension of the null space of $A$. What is the rank of $A$?
   
   Solution: Since row operations do not affect the dependence relations of the columns, and the first, second, and fourth columns of the RREF matrix are independent, the same columns of $A$ must be independent and form a spanning set for $Col\;A$. So
   $$Col\;A=span\left\{\left(\begin{array}{r} 1\\ 1\\ 1\\ 1\\ \en... ...\right),\left(\begin{array}{r} 3\\ 6\\ 10\\ 7\\ \end{array}\right)\right\}$$
   and $Col\;A$ has dimension 3, and so the rank is also three. Then since $dim(null\;A)+dim(Col\;A)=5$ the null space must have dimension 2.
   

3. Determine which of the following sets of vectors form a basis for $\mathbb{R}^3$. If a set is not a basis, determine the dimension of the subspace spanned by the set.
   

   1. $$\left(\begin{array}{c} 1\\ 0\\ 0\\ \end{array}\right),\left(\... ... \end{array}\right),\left(\begin{array}{c} 1\\ 1\\ 1\\ \end{array}\right)$$
      
      Solution:Three linearly independent vectors in $\mathbb{R}^3$ always form a basis for $\mathbb{R}^3$. So this set forms a basis.
      

   2. $$\left(\begin{array}{c} 1\\ 0\\ 1\\ \end{array}\right),\left(\... ... \end{array}\right),\left(\begin{array}{c} 2\\ 4\\ 1\\ \end{array}\right)$$
      
      Solution:Any set of vectors with the zero vector is dependent, so this can not be a basis for $\mathbb{R}^3$. If we place these vectors into a matrix and row reduce we will find that the dimension of the column space of this matrix is two and so the vectors span a 2-dimensional subspace of $\mathbb{R}^3$.
      

   3. $$\left(\begin{array}{r} 1\\ -4\\ 3\\ \end{array}\right),\left(... ... \end{array}\right),\left(\begin{array}{r} 0\\ 2\\ -2\\ \end{array}\right)$$
      
      Solution:No set of four vectors in $\mathbb{R}^3$ can be independent, so this set is not a basis for $\mathbb{R}^3$. If we place these vectors into a matrix and row reduce we find that
      $$\left(\begin{array}{rrrr} 1&0&3&0\\ -4&3&-5&2\\ 3&-1&4&-2\\ ... ...begin{array}{cccc} 1&0&0&0\\ 0&1&0&-1/2\\ 0&0&1&1/2\\ \end{array}\right),$$
      and the dimension of the column space is three and so the vectors span a 3-dimnensional subspace of $\mathbb{R}^3$, and so the vectors span all of $\mathbb{R}^3$. So it is possible to have a spanning set which is not a basis.
      

   4. $$\left(\begin{array}{r} 1\\ 0\\ 2\\ \end{array}\right),\left(\begin{array}{c} 0\\ 3\\ -1\\ \end{array}\right)$$
      
      Solution:Two independent vectors in $\mathbb{R}^3$ span a 2-dimensional subspace of $\mathbb{R}^3$ and so are not a basis for $\mathbb{R}^3$.
      

   5. $$\left(\begin{array}{r} 1\\ -1\\ 2\\ \end{array}\right),\left(\begin{array}{c} -3\\ 3\\ -6\\ \end{array}\right)$$
      
      Solution:Two dependent vectors in $\mathbb{R}^3$ span a 1-dimensional subspace of $\mathbb{R}^3$ and so are not a basis for $\mathbb{R}^3$.
      

4. Find a basis for the subspace spanned by the following sets of vectors.

   1. $$\left(\begin{array}{r} 1\\ 0\\ 0\\ 1\\ \end{array}\right),\l... ...d{array}\right),\left(\begin{array}{r} 0\\ 3\\ -1\\ 1\\ \end{array}\right)$$
      
      Solution:Placing these vectors into a matrix we get
      $$\left(\begin{array}{rrrrr} 1&-2&6&5&0\\ 0&1&-1&-2&3\\ 0&-1&2&... ...rr} 1&0&0&0&-2\\ 0&1&0&0&5\\ 0&0&1&0&2\\ 0&0&0&1&0\\ \end{array}\right)$$
      and so these vectors span a 4-dimensional subspace of $\mathbb{R}^4$, which is all of $\mathbb{R}^4$, and so in this instance any basis of $\mathbb{R}^4$ will do.
      $$\left\{\left(\begin{array}{r} 1\\ 0\\ 0\\ 0\\ \end{array}\ri... ...}\right),\left(\begin{array}{r} 0\\ 0\\ 0\\ 1\\ \end{array}\right)\right\}$$

   2. $$\left(\begin{array}{r} 1\\ -3\\ -2\\ \end{array}\right),\left... ... \end{array}\right),\left(\begin{array}{r} 2\\ -6\\ -4\\ \end{array}\right)$$
      
      Solution:Since
      $$\left(\begin{array}{rrr} 1&3&2\\ -3&-9&-6\\ -2&-6&-4\\ \end... ...t)\sim\left(\begin{array}{rrr} 1&3&2\\ 0&0&0\\ 0&0&0\\ \end{array}\right)$$
      these vectors span a 1-dimensional subspace with the following basis
      $$\left\{\left(\begin{array}{r} 1\\ -3\\ -2\\ \end{array}\right)\right\}$$

5. Let $T:\mathbb{R}^5\mapsto \mathbb{R}^4$ be a linear transformation defined as
   $$T\left(\begin{array}{r} x\\ y\\ z\\ w\\ t\\ \end{array}\right)=\left(\begin{array}{r} y\\ w\\ z\\ x\\ \end{array}\right)$$
   Apply $T$ to each vector in the standard basis for $\mathbb{R}^5$, and then decide if a linear transformation always transforms a linearly independent set to another linearly indepenent set.
   
   Solution:Since the set $\{T(\bar{e}_1),T(\bar{e}_2),T(\bar{e}_3),T(\bar{e}_4),T(\bar{e}_5)\}= \{\bar{e}_4,\bar{e}_1,\bar{e}_3,\bar{e}_2,\bar{0}\}$ the set of images is dependent in $\mathbb{R}^4$ (since the zero vector is in the set) and a linear map can take an independent set to a dependent set.
   

6. Suppose that $T:V\mapsto W$ is a ono-to-one linear transformation; i.e., if $T(\v )=T(\u )$ then $\v =\u$. Show that if the set of images $\{T(\v _1),T(v_2),\dots,T(\v _p)\}$ is a linearly dependent set, the the set $\{\v _1,\v _2,\dots,\v _p\}$ is also a linearly dependent set. This shows that a one-to-one linear transformation always maps a linearly independent set to another linearly independent set.
   
   Solution:Suppose that there is some non-trivial linear combination
   $$c_1T(\v _1)+c_2T(v_2)+\cdots+c_pT(\v _p)=0$$
   Since $T$ is linear this gives us that
   $$T(c_1\v _1+c_2\v _2+\cdots+c_p\v _p)=0$$
   But since $T$ is one-to-one, $T(\bar{x})=\bar{0}$ has only the trivial solution. So we have a non-trivial linear combination
   $$c_1\v _1+c_2\v _2+\cdots+c_p\v _p=0$$
   and so $\{\v _1,\v _2,\dots,\v _p\}$ is a linearly dependent set.
   

7. Find the change-of-coordinates matrix that takes the basis
   $$\mathcal{B}_1=\left\{\left(\begin{array}{c} 3\\ -1\\ 4\\ \end... ...rray}\right),\left(\begin{array}{c} 8\\ -2\\ 7\\ \end{array}\right)\right\}$$
   to the standard basis $\mathcal{B}_S$. Hint: What is the change-of-cordinates matrix that mapped the standard basis $\mathcal{B}_S$ to $\mathcal{B}_1$?
   
   Solution:The matrix to change from the standard coordinates to the non-standard coordinates is simply the inverse of the matrix with the non-standard basis as its columns.
   $$\left(\begin{array}{ccc} 3&2&8\\ -1&0&-2\\ 4&-5&7\\ \end{ar... ...cc} -5/4&-27/4&-1/2\\ -1/8&-11/8&-1/4\\ 5/8&23/8&1/4\\ \end{array}\right)$$
   This is the change-of-coordinates matrix. However, to change from the non-standard coordinates back to standard coordinates we use the matrix that has the non-standard basis as its columns. So we would use
   $$\left(\begin{array}{ccc} 3&2&8\\ -1&0&-2\\ 4&-5&7\\ \end{array}\right)$$

8. Given the basis $\mathcal{B}_1$ from the previous problem, give the standard basis coordinates of $\v$ (that is write $[\v ]_{\mathcal{B}_S}$) when
   $$\v =\left[\begin{array}{c} 3\\ -1\\ 2\\ \end{array}\right]_{\mathcal{B}_1}$$
   
   Solution:

   
   

   $$[\v ]_{\mathcal{B}_S} = 3\left(\begin{array}{c} 3\\ -1\\ 4\\ \end{array}\right)-\left(\b... ...\\ \end{array}\right)+2\left(\begin{array}{c} 8\\ -2\\ 7\\ \end{array}\right)$$

   $$= \left(\begin{array}{r} 23\\ -7\\ 21\\ \end{array}\right)$$

   
   

9. Cramer's rule (which we will define below) is used in a variety of theoretical calculations, especially when examining how the solution of $A\bar{x}=\b$ is affected when a single entry of $\b$ is changed.
   

   For any $n\times n$ matrix $A=[\bar{a}_1\;\;\bar{a}_2\;\;\cdots\;\;\bar{a}_{n-1}\;\;\bar{a}_n]$ and any $\b\in\mathbb{R}^n$, define $A_i(b)$ as the $n\times n$ matrix obtained by replacing the $i^{th}$ column of $A$ with the vector $\b$.

   $$A_i(b)=[\bar{a}_1\;\;\cdots\;\;\bar{a}_{i-1}\;\;\b\;\;\; \bar{a}_{i+1}\;\;\cdots\;\;\bar{a}_{n}]$$
   Theorem 1 (Cramer's Rule)   Let $A$ be an invertible $n\times n$ matrix. Then for any $\b\in\mathbb{R}^n$, the unique solution $\bar{x}$ of $A\bar{x}=\b$ has entries given by

   $$x_i=\displaystyle \frac{det\;A_i(b)}{det\;A},\;\;\;i=1,2,\dots,n$$

   
   

   Use Cramer's rule to find the solution to the linear system
   

   $$2x_1+x_2+x_3 = 4$$

   $$-x_1+2x_3 = 2$$

   $$3x_1+x_2+3x_3 = -2$$

   
   
   
   Solution:The coefficient matrix is
   $$A=\left(\begin{array}{ccc} 2&1&1\\ -1&0&2\\ 3&1&3\\ \end{array}\right)$$
   We first compute

   
   

   $$\left\vert\begin{array}{ccc} 2&1&1\\ -1&0&2\\ 3&1&3\\ \end{array}\right\vert = -1\left\vert\begin{array}{rr} -1&2\\ 3&3\\ \end{array}\right\vert-1\left\vert\begin{array}{rr} 2&1\\ -1&2\\ \end{array}\right\vert$$

   $$= (-1)(-9)+(-1)(5)$$

   $$= 4$$

   
   

   So

   
   

   $$x_1=\displaystyle \frac{det\;A_1(\b )}{det\;A} = \displaystyle \frac{\left\vert\begin{array}{ccc} 4&1&1\\ 2&0&2\\ -2&1&3\\ \end{array}\right\vert}{4}$$

   $$= \displaystyle \frac{-1\left\vert\begin{array}{rr} 2&2\\ -2&3\\ ... ...t\vert+(-1)\left\vert\begin{array}{rr} 4&1\\ 2&2\\ \end{array}\right\vert}{4}$$

   $$= \displaystyle \frac{(-1)(10)+(-1)(6)}{4}$$

   $$= -4$$

   
   

   
   

   $$x_2=\displaystyle \frac{det\;A_2(\b )}{det\;A} = \displaystyle \frac{\left\vert\begin{array}{ccc} 2&4&1\\ -1&2&2\\ 3&-2&3\\ \end{array}\right\vert}{4}$$

   $$= \displaystyle \frac{2\left\vert\begin{array}{rr} 2&2\\ -2&3\\ \... ...ight\vert+3\left\vert\begin{array}{rr} 4&1\\ 2&2\\ \end{array}\right\vert}{4}$$

   $$= \displaystyle \frac{2(10)+(14)+3(6)}{4}$$

   $$= 13$$

   
   

   
   

   $$x_3=\displaystyle \frac{det\;A_3(\b )}{det\;A} = \displaystyle \frac{\left\vert\begin{array}{ccc} 2&1&4\\ -1&0&2\\ 3&1&-2\\ \end{array}\right\vert}{4}$$

   $$= \displaystyle \frac{-1\left\vert\begin{array}{rr} -1&2\\ 3&-2\\ ... ...ght\vert-1\left\vert\begin{array}{rr} 2&4\\ -1&2\\ \end{array}\right\vert}{4}$$

   $$= \displaystyle \frac{(-1)(-4)+(-1)(8)}{4}$$

   $$= -1$$

   
   

   So

   $$\bar{x}=\left(\begin{array}{c} -4\\ 13\\ -1\\ \end{array}\right)$$
   A quick check shows that
   $$\left(\begin{array}{ccc} 2&1&1\\ -1&0&2\\ 3&1&3\\ \end{arra... ... \end{array}\right)=\left(\begin{array}{c} 4\\ 2\\ -2\\ \end{array}\right)$$

10. Let $T:\P _3(\mathbb{R})\mapsto \mathbb{R}^4$ be the linear transformation defined so that if $\bar{p}\in\P _3(\mathbb{R})$, then
    $$T(\bar{p})=\left(\begin{array}{c} p(0)\\ p(2)\\ p(1)\\ p(7)\\ \end{array}\right)$$
    Describe the kernel of $T$. Hint: The Fundamental Theorem of Algebra states that any non-zero polynomial of degree $n$ can have no more than $n$ roots.
    
    Solution:Any polynomial in the kernel of $T$ must have the following roots; $x=0$, $x=2$, $x=1$, $x=7$. But any polynomial in $\P _3(\mathbb{R})$ has degree less than four and so the only polynomial in the kernel is the zero polynomial.
    

11. Show that the set $\{x^2-x,\;2x-1,\;1-x-x^2\}$ does NOT form a basis for $\P _2(\mathbb{R})$. That is, find a non-trivial linear combination of these polynomials that equals the zero polynomial.
    
    Solution:If we have all weights of the combination equal to one we get
    $$x^2-x+2x-1+1-x-x^2=0$$
    which is a non-trivial linear combination producing the zero vector. The set is therefore dependent.
    

12. Find a basis for $\mathcal{M}_{3\times 3}(\mathbb{R})=\left\{\left(\begin{array}{ccc} a&b&c\\ d&e&f\\ g&h&i\\ \end{array}\right):a,b,c,d,e,f,g,h,i\in\mathbb{R}\right\}$.
    
    What is the dimension of this vector space? In general, what will be the dimension of $\mathcal{M}_{m\times n}$, which is the vector space of all $m\times n$ matrices?
    
    Solution:Every matrix in $\mathcal{M}_{3\times 3}$ can be written as a linaer combination of the following set of 9 matrices.
    $$\left(\begin{array}{ccc} 1&0&0\\ 0&0&0\\ 0&0&0\\ \end{array... ...ight),\left(\begin{array}{ccc} 0&0&1\\ 0&0&0\\ 0&0&0\\ \end{array}\right)$$
    $$\left(\begin{array}{ccc} 0&0&0\\ 1&0&0\\ 0&0&0\\ \end{array... ...ight),\left(\begin{array}{ccc} 0&0&0\\ 0&0&1\\ 0&0&0\\ \end{array}\right)$$
    $$\left(\begin{array}{ccc} 0&0&0\\ 0&0&0\\ 1&0&0\\ \end{array... ...ight),\left(\begin{array}{ccc} 0&0&0\\ 0&0&0\\ 0&0&1\\ \end{array}\right)$$
    Since this set is linearly independent it forms a basis for $\mathcal{M}_{3\times 3}$ and so $\mathcal{M}_{3\times 3}$ is a 9-dimensional vector space. In general, $\mathcal{M}_{m\times n}$ will be an $mn$-dimensional vector space.
    

13. Recall that a $n\times n$ matrix $A$ is called nilpotent of class $k$ if $A^k=0$ and $A^{k-1}\not=0$ for some positive integer $k$. Suppose an $n\times n$ matrix $A$ has nilpotency class $k$. Then there will be some vector $\v\in\ensuremath{\mathbb{R}^n}$ such that $A^{k-1}\v\not=\bar{0}$. It is obvious that no vector from the set $\left\{\v ,A\v ,A^2\v ,\dots,A^{k-1}\v\right\}$ is the zero vector. Show that the set of vectors $\left\{\v ,A\v ,A^2\v ,\dots,A^{k-1}\v\right\}$ is a basis for a $k$-dimensional subspace of $\ensuremath{\mathbb{R}^n}$.
    

    Hint: Consider a linear combination

    $$c_0\v +c_1A\v +c_sA^2\v +\cdots+c_{k-1}A^{k-1}\v =\bar{0}$$
    and multiply both sides by $A^{k-1}$ to show that $c_0=0$. Next show that $c_1=0$, and so on.
    
    Solution:Using the hint we get
    

    $$A^{k-1}(c_0\v +c_1A\v +c_sA^2\v +\cdots+c_{k-1}A^{k-1}\v ) = A^{k-1}\bar{0}$$

    $$c_0A^{k-1}\v +c_1A^k\v +c_sA^{k+2}\v +\cdots+c_{k-1}A^{2k-2}\v = \bar{0}$$

    $$c_0A^{k-1}\v = \bar{0}$$

    
    

    Since $A^{k-1}\v\not=0$, $c_0=0$. Now multiply the resulting linear combination

    $$0\cdot\v +c_1A\v +c_sA^2\v +\cdots+c_{k-1}A^{k-1}\v =\bar{0}$$
    by $A^{k-2}$ to get

    
    

    $$A^{k-1}(0\cdot\v +c_1A\v +c_sA^2\v +\cdots+c_{k-1}A^{k-1}\v ) = A^{k-2}\bar{0}$$

    $$c_1A^{k-1}\v +c_sA^{k+1}\v +\cdots+c_{k-1}A^{2k-3}\v = \bar{0}$$

    $$c_0A^{k-1}\v = \bar{0}$$

    
    
    and the same argument shows that $c_1=0$. Continue in the same fashion, decreasing the exponent over $A$ by one in each step to see that every weight $c_i=0$. Therefore, by definition, the set is linearly independent.