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Prove that $V = textnull varphi oplus au:a in mathbbF$

Prove that if $(v_1,ldots,v_n)$ spans $V$, then so does the list $(v_1-v_2,v_2-v_3,ldots,v_n-1-v_n,v_n).$Questions about if $dim(U)gedim(V)−dim(W)$ and proving $∃T∈mathfrakL(V,W) texts.t.textnull(T)=U$?Is this sufficient for linear independence proofs??Proof that the span of a list is equal to the span of any reordering of the listSuppose $P in mathscrL(V)$ and $P^2 = P$. Prove that $V = textnullP oplus textrange P$There exists a subspace $U$ of $V$ such that $Ucap null T=0$ and $range T = mathcalJ=Tu$There exists a linear transformation $S$ such that $V = N_T oplus R_S$A Change of Basis.Prove that $textnull(T_1)= textnull(T_2)$ if f $exists$ $S in L(W,W) $ such that $T_2 = ST_1$Is the linear map on basis of $V$ a basis of $W$?

1

$begingroup$

As I am self learning Linear Algebra Done Right, I would like to make sure I learn it correctly. Please help on the following the proof.

Suppose $varphi in mathcalL(V, mathbbF)$. Suupose $u in V$ is not in null $varphi$. Prove that $V = textnull varphi oplus au:a in mathbbF$

My approach:

Suppose $varphi in mathcalL(V, mathbbF)$, and $u in V$ is not in null $varphi$. Let $U = au:a in mathbbF$ .

Let $w_1,...,w_m$ be the basis of null $varphi$. Extend $w_1,...,w_m$ to be a basis of $V$ as $w_1,...,w_m, v_1,...,v_n$. Then $v_1,...,v_n$ is the basis of $au:a in mathbbF$ [I cannot convince myself here. Seems right.Not sure]

For any $v in V$, $v = a_1m_1 + ... + a_mw_m + b_1v_1+...+b_nv_n = textnull varphi + U$.

Suppose $v in textnull varphi cap U$, then $v = a_1m_1 + ... + a_mw_m = b_1v_1+...+b_nv_n $. $a_1m_1 + ... + a_mw_m - b_1v_1 -...-b_nv_n = 0$. Since $w_1,...,w_m, v_1,...,v_n$ is a basis of $V$, $a_1 = ... = a_m = b_1 = ... = b_n = 0$ which implies $v = 0$.

Therefore $V = textnull varphi oplus au:a in mathbbF$.

linear-algebra proof-verification linear-transformations

share|cite|improve this question

asked Mar 13 at 10:14

JOHN JOHN

4279

$endgroup$

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  $begingroup$
  The set $ au : ain BbbF $ is just $span(u)$. Since $u$ is nonzero (otherwise it would have to be mapped to zero under $varphi$ and would hence be in the null space) $U$ is a one-dimensional subspace with $u$ itself as a basis.
  $endgroup$
  – Alex Sanger
  Mar 13 at 15:40
  
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  There is no assumption in this exercise that V is finite-dimensional. Thus you should stay away from bases.
  $endgroup$
  – Sheldon Axler
  Mar 14 at 0:41
  

  
  
  
  

add a comment | 

1

$begingroup$

As I am self learning Linear Algebra Done Right, I would like to make sure I learn it correctly. Please help on the following the proof.

Suppose $varphi in mathcalL(V, mathbbF)$. Suupose $u in V$ is not in null $varphi$. Prove that $V = textnull varphi oplus au:a in mathbbF$

My approach:

Suppose $varphi in mathcalL(V, mathbbF)$, and $u in V$ is not in null $varphi$. Let $U = au:a in mathbbF$ .

Let $w_1,...,w_m$ be the basis of null $varphi$. Extend $w_1,...,w_m$ to be a basis of $V$ as $w_1,...,w_m, v_1,...,v_n$. Then $v_1,...,v_n$ is the basis of $au:a in mathbbF$ [I cannot convince myself here. Seems right.Not sure]

For any $v in V$, $v = a_1m_1 + ... + a_mw_m + b_1v_1+...+b_nv_n = textnull varphi + U$.

Suppose $v in textnull varphi cap U$, then $v = a_1m_1 + ... + a_mw_m = b_1v_1+...+b_nv_n $. $a_1m_1 + ... + a_mw_m - b_1v_1 -...-b_nv_n = 0$. Since $w_1,...,w_m, v_1,...,v_n$ is a basis of $V$, $a_1 = ... = a_m = b_1 = ... = b_n = 0$ which implies $v = 0$.

Therefore $V = textnull varphi oplus au:a in mathbbF$.

linear-algebra proof-verification linear-transformations

share|cite|improve this question

asked Mar 13 at 10:14

JOHN JOHN

4279

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The set $ au : ain BbbF $ is just $span(u)$. Since $u$ is nonzero (otherwise it would have to be mapped to zero under $varphi$ and would hence be in the null space) $U$ is a one-dimensional subspace with $u$ itself as a basis.
  $endgroup$
  – Alex Sanger
  Mar 13 at 15:40
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  There is no assumption in this exercise that V is finite-dimensional. Thus you should stay away from bases.
  $endgroup$
  – Sheldon Axler
  Mar 14 at 0:41
  

  
  
  
  

add a comment | 

$begingroup$

As I am self learning Linear Algebra Done Right, I would like to make sure I learn it correctly. Please help on the following the proof.

Suppose $varphi in mathcalL(V, mathbbF)$. Suupose $u in V$ is not in null $varphi$. Prove that $V = textnull varphi oplus au:a in mathbbF$

My approach:

Suppose $varphi in mathcalL(V, mathbbF)$, and $u in V$ is not in null $varphi$. Let $U = au:a in mathbbF$ .

Let $w_1,...,w_m$ be the basis of null $varphi$. Extend $w_1,...,w_m$ to be a basis of $V$ as $w_1,...,w_m, v_1,...,v_n$. Then $v_1,...,v_n$ is the basis of $au:a in mathbbF$ [I cannot convince myself here. Seems right.Not sure]

For any $v in V$, $v = a_1m_1 + ... + a_mw_m + b_1v_1+...+b_nv_n = textnull varphi + U$.

Suppose $v in textnull varphi cap U$, then $v = a_1m_1 + ... + a_mw_m = b_1v_1+...+b_nv_n $. $a_1m_1 + ... + a_mw_m - b_1v_1 -...-b_nv_n = 0$. Since $w_1,...,w_m, v_1,...,v_n$ is a basis of $V$, $a_1 = ... = a_m = b_1 = ... = b_n = 0$ which implies $v = 0$.

Therefore $V = textnull varphi oplus au:a in mathbbF$.

linear-algebra proof-verification linear-transformations

share|cite|improve this question

asked Mar 13 at 10:14

JOHN JOHN

4279

$endgroup$

share|cite|improve this question

asked Mar 13 at 10:14

JOHN JOHN

4279

share|cite|improve this question

asked Mar 13 at 10:14

JOHN JOHN

4279

asked Mar 13 at 10:14

JOHN JOHN

4279

asked Mar 13 at 10:14

JOHN JOHN

4279