<p>Your formula</p>
<p><span class="math-container">$$N(A) \oplus  R(A) = V\qquad (1)$$</span></p>
<p>is not true in general cases, it is correct  in a particular case, if <span class="math-container">$A$</span> is a symmetric matrix, so that <span class="math-container">$A^T=A$</span>, where <span class="math-container">$A^T$</span> is the transpose of <span class="math-container">$A$</span>.</p>
<p>Actually, there is in linear algebra a theorem that involves the direct sum of <em>kernel</em> and <em>range</em> of a linear transformation, and involves the transpose <span class="math-container">$A^T$</span> of the matrix <span class="math-container">$A$</span>. And this theorem holds for any real matrix, not only for square matrices.</p>
<p>If you don't mind,  I change the notations, calling <span class="math-container">$\ker A$</span> (or  <span class="math-container">$\ker L_A$</span>) the <em>kernel</em> (or <em>null space</em>) of the linear tranformation associated to <span class="math-container">$A$</span> (your <span class="math-container">$N(A)$</span>) and <span class="math-container">$\operatorname{Im} A$</span> ( or <span class="math-container">$\operatorname{Im} L_A$</span>) your <span class="math-container">$R(A)$</span>, the <em>image</em> or <em>range</em> of the linear transformation associated to <span class="math-container">$A$</span> (I prefer this notation to not get confused, as I find easier to remember).</p>
<p>The theorem I referred to above says:</p>
<p><em><strong>Theorem.</strong></em> Let be <span class="math-container">$A \in M_{m,n}(\mathbb{R})$</span>, that is a matrix <span class="math-container">$m\times n$</span> with real entries. Then</p>
<p>(i) <span class="math-container">$\mathbb{R^m}= \operatorname{Im}  L_A\oplus \ker L_{A^T}$</span> and  <span class="math-container">$\mathbb{R^n}=  \operatorname{Im}L_{A^T} \oplus \ker L_A$</span>,</p>
<p>(ii) <span class="math-container">$\operatorname{rank} A= \operatorname{rank} A^T$</span>,</p>
<p>where <span class="math-container">$L_A$</span> is the linear transformation associated to matrix <span class="math-container">$A$</span>.</p>
<p><span class="math-container">$\;$</span></p>
<p>The second statement (ii) is an important result of linear algebra, which says that the <strong>rank by columns of a matrix is equal to its rank by rows,</strong> that is the maximum number of linearly indipendent columns is equal to the maximum number of linearly indipendent rows.</p>
<p>I omit the proof I know, as it is too complicated and cumbersome as it recalls theorems and proposition previously stated, and it is necessary to recall and restate them. And in general the proofs and the order of theorems and definitions can change among books.</p>
<p>So, it is better you look at this theorem on your book or some other book of linear algebra, also because, as I said, this is an important result.</p>
<p>Anyway, <em>Grassman formula</em> and the  <em>Rank-nullity</em> theorem in your question  are used in the proof.</p>

CC BY-SA 4.0
2023-03-15T19:51:20.087Z
2023-03-15T22:00:28.057Z
2023-03-15T22:00:28.057Z
BakerStreet
<p>If <span class="math-container">$A$</span> is a <span class="math-container">$m•n$</span> matrix.</p>
<p><strong>Question:</strong> Is <span class="math-container">$N(A) \oplus  R(A) = V \; \forall A$</span>  ?</p>
<ul>
<li><em>Update: I now think this question makes sense only for square matrices, as noted below.</em></li>
</ul>
<p><strong>Terminology</strong></p>
<ol>
<li><span class="math-container">$R(A)$</span> By this I mean the Range of Matrix A, which is also defined as <span class="math-container">$CS(A)$</span> the column space of <span class="math-container">$A$</span>.</li>
<li><span class="math-container">$N(A)$</span> the Null Space of Matrix <span class="math-container">$A$</span>.</li>
<li><span class="math-container">$\oplus$</span> Direct Sum as defined in point 3 below.</li>
<li>I assume <span class="math-container">$V$</span> to be <span class="math-container">$\mathbb{R^n}$</span></li>
</ol>
<p><em>Why am I posting here? Linear Algebra is fundamental to alot of economic study, and I find this community much more patient, engaging, and generous than the community at StackMaths...(don't tell anyone!)</em></p>
<p><strong>My thoughts so far:</strong></p>
<ol start="0">
<li>Thanks to Michael Greinecker's comment, I believe that we must have <span class="math-container">$n=m$</span> otherwise we cannot take the sum <span class="math-container">$R(A) + N(A)$</span> because they would be subspaces of different dimensions....But still the question holds for when <span class="math-container">$n=m$</span>. Thanks!</li>
<li>I know that If A is a projection Matrix P then <span class="math-container">$N(P) \oplus  R(P) = V \; \forall P$</span>. But what about any Square <span class="math-container">$A$</span>?</li>
<li>I understand that if we have a symmetric matrix, then <span class="math-container">$N(A) \oplus  R(A) = V$</span> as the <span class="math-container">$N(A) = N(A^T) = R(A)^\perp$</span> and the sum of orthogonal complements is direct.</li>
<li><span class="math-container">$V = N(A) \oplus  R(A)$</span> means that  <span class="math-container">$V = R(A) + N(A)$</span> &amp; <span class="math-container">$N(A) \cap R(A) = \{0\}$</span></li>
</ol>
<ul>
<li>Using the Rank Nullity Theorem <span class="math-container">$Dim(V) = Dim(N(A)) + Dim(Range(A))$</span></li>
<li>I believe <span class="math-container">$R(A)$</span> &amp; <span class="math-container">$N(A)$</span> are always Linearly Independent?</li>
</ul>
<p>Therefore we would have <span class="math-container">$V = R(A) + N(A)$</span> &amp; as they are both subspaces <span class="math-container">$N(A) \cap R(A) = \{0\}$</span>.</p>
<p>If they were not Linearly Independent then we must surely have <span class="math-container">$N(A) \subseteq R(A)$</span> or vice versa.</p>
<p>I can't figure out what this would mean, or if this every happens except the trivial case where A is non-singular (in which case <span class="math-container">$N(A) \oplus  R(A) = V$</span> still holds).</p>
<ul>
<li>As a corollary of what I'm saying, is it correct that our vector spaces either only share the 0 vector, or one is a subset of the other. I.e. they can't just share some vectors.</li>
</ul>

CC BY-SA 4.0
2023-03-14T21:17:23.520Z
Hi! Did you mean to post this here, and not at [Math.se]?
Giskard
CC BY-SA 4.0
2023-03-15T00:08:36.973Z
What is $R(A)$? If it is the range of the corresponding mapping, then there is no reason that it should be a subspace of $V$, which is presumably the domain. Generally, it would be helpful to explain notation and definitions. Different equivalent definitions can require different proof approaches.
Michael Greinecker
CC BY-SA 4.0
2023-03-15T00:36:45.140Z
Hi @Giskard Yes I did! I hope you don't mind, I briefly outlined why in the post. But basically I prefer you guys lol, and my degree is Mathematics and Economics...so it seems legit. (Plus all this is essential for econometrics so I'm sure people here know).
CormJack
CC BY-SA 4.0
2023-03-15T00:40:17.413Z
@MichaelGreinecker thanks for this! I have trued to clarify in the question now. And yes now I have been more clear I think it's valid to say "For any m•n Matrix A, R(A) is a subspace of $\mathbb{R^m}$" This is a quote from Martin Anthony's Linear Algebra book. And similar N(A) is a subspace of $\mathbb{R^n}$  apologies for my initial lack of clarity, let me know if I've missed anything else, thanks!
CormJack
CC BY-SA 4.0
2023-03-15T20:12:38.740Z
By the way, what do you mean when you say" $R(A)$ & $N(A)$ are always Linearly Independent?". Linear indipendence is a concept relative to  sets of vectors, not subspaces.
BakerStreet
CC BY-SA 4.0
2023-03-15T20:39:15.010Z
I edited your formula of the Rank-Nullity Theorem as I think there was a typo, it is dim (Range) not dim (Rank).
BakerStreet
<mathematical-economics><linear-algebra>
$N(A) \oplus R(A) = V \; \forall A$?
CormJack
# $N(A) \oplus R(A) = V \; \forall A$?

(Asked by: username_0 on 2023-03-14)

If $A$ is a $m•n$ matrix.

**Question:** Is $N(A) \oplus R(A) = V \; \forall A$ ?

* *Update: I now think this question makes sense only for square matrices, as noted below.*

**Terminology**

1. $R(A)$ By this I mean the Range of Matrix A, which is also defined as $CS(A)$ the column space of $A$.
2. $N(A)$ the Null Space of Matrix $A$.
3. $\oplus$ Direct Sum as defined in point 3 below.
4. I assume $V$ to be $\mathbb{R^n}$

*Why am I posting here? Linear Algebra is fundamental to alot of economic study, and I find this community much more patient, engaging, and generous than the community at StackMaths...(don't tell anyone!)*

**My thoughts so far:**

0. Thanks to Michael Greinecker's comment, I believe that we must have $n=m$ otherwise we cannot take the sum $R(A) + N(A)$ because they would be subspaces of different dimensions....But still the question holds for when $n=m$. Thanks!
1. I know that If A is a projection Matrix P then $N(P) \oplus R(P) = V \; \forall P$. But what about any Square $A$?
2. I understand that if we have a symmetric matrix, then $N(A) \oplus R(A) = V$ as the $N(A) = N(A^T) = R(A)^\perp$ and the sum of orthogonal complements is direct.
3. $V = N(A) \oplus R(A)$ means that $V = R(A) + N(A)$ & $N(A) \cap R(A) = \{0\}$

* Using the Rank Nullity Theorem $Dim(V) = Dim(N(A)) + Dim(Range(A))$
* I believe $R(A)$ & $N(A)$ are always Linearly Independent?

Therefore we would have $V = R(A) + N(A)$ & as they are both subspaces $N(A) \cap R(A) = \{0\}$.

If they were not Linearly Independent then we must surely have $N(A) \subseteq R(A)$ or vice versa.

I can't figure out what this would mean, or if this every happens except the trivial case where A is non-singular (in which case $N(A) \oplus R(A) = V$ still holds).

* As a corollary of what I'm saying, is it correct that our vector spaces either only share the 0 vector, or one is a subset of the other. I.e. they can't just share some vectors.

**Tags:** mathematical-economics, linear-algebra
> **Comments:**
> **username_1**: Hi! Did you mean to post this here, and not at [Math.se]?  
> **username_2**: What is $R(A)$? If it is the range of the corresponding mapping, then there is no reason that it should be a subspace of $V$, which is presumably the domain. Generally, it would be helpful to explain notation and definitions. Different equivalent definitions can require different proof approaches.  
> **username_0**: Hi @Giskard Yes I did! I hope you don't mind, I briefly outlined why in the post. But basically I prefer you guys lol, and my degree is Mathematics and Economics...so it seems legit. (Plus all this is essential for econometrics so I'm sure people here know).  
> **username_0**: @MichaelGreinecker thanks for this! I have trued to clarify in the question now. And yes now I have been more clear I think it's valid to say "For any m•n Matrix A, R(A) is a subspace of $\mathbb{R^m}$" This is a quote from Martin Anthony's Linear Algebra book. And similar N(A) is a subspace of $\mathbb{R^n}$  apologies for my initial lack of clarity, let me know if I've missed anything else, thanks!  
> **username_3**: By the way, what do you mean when you say" $R(A)$ & $N(A)$ are always Linearly Independent?". Linear indipendence is a concept relative to  sets of vectors, not subspaces.  
> **username_3**: I edited your formula of the Rank-Nullity Theorem as I think there was a typo, it is dim (Range) not dim (Rank).  

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## 1 answers

### Answer by username_3 on 2023-03-15. Score: 3 | Accepted answer

Your formula

$$N(A) \oplus R(A) = V\qquad (1)$$

is not true in general cases, it is correct in a particular case, if $A$ is a symmetric matrix, so that $A^T=A$, where $A^T$ is the transpose of $A$.

Actually, there is in linear algebra a theorem that involves the direct sum of *kernel* and *range* of a linear transformation, and involves the transpose $A^T$ of the matrix $A$. And this theorem holds for any real matrix, not only for square matrices.

If you don't mind, I change the notations, calling $\ker A$ (or $\ker L\_A$) the *kernel* (or *null space*) of the linear tranformation associated to $A$ (your $N(A)$) and $\operatorname{Im} A$ ( or $\operatorname{Im} L\_A$) your $R(A)$, the *image* or *range* of the linear transformation associated to $A$ (I prefer this notation to not get confused, as I find easier to remember).

The theorem I referred to above says:

***Theorem.*** Let be $A \in M\_{m,n}(\mathbb{R})$, that is a matrix $m\times n$ with real entries. Then

(i) $\mathbb{R^m}= \operatorname{Im} L\_A\oplus \ker L\_{A^T}$ and $\mathbb{R^n}= \operatorname{Im}L\_{A^T} \oplus \ker L\_A$,

(ii) $\operatorname{rank} A= \operatorname{rank} A^T$,

where $L\_A$ is the linear transformation associated to matrix $A$.

$\;$

The second statement (ii) is an important result of linear algebra, which says that the **rank by columns of a matrix is equal to its rank by rows,** that is the maximum number of linearly indipendent columns is equal to the maximum number of linearly indipendent rows.

I omit the proof I know, as it is too complicated and cumbersome as it recalls theorems and proposition previously stated, and it is necessary to recall and restate them. And in general the proofs and the order of theorems and definitions can change among books.

So, it is better you look at this theorem on your book or some other book of linear algebra, also because, as I said, this is an important result.

Anyway, *Grassman formula* and the *Rank-nullity* theorem in your question are used in the proof.


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