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Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}. $$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$ A(e^{tX},1)\in \Mat_n^{1,2,3}. $$ Thus

$$ Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$ X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}. $$ Thus $X^2\oplus0\in\Mat_n^{1,2}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=-\bu_1,\;\;X^2\bu_2=-\bu_2,\;\;X^2\bu_3=0. $$

Now consider the matrix

$$Y_{\bu_1}=-\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big). $$

Note that

$$ Y_{\bu_1}\bu_1=\bu_1,\;\;Y_{\bu_1}\bu_2=Y_{\bu_1}\bu_3=0. $$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i. $$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.

Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}. $$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$ A(e^{tX},1)\in \Mat_n^{1,2,3}. $$ Thus

$$ Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$ X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}. $$ Thus $X^2\oplus0\in\Mat_n^{1,2,3}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=-\bu_1,\;\;X^2\bu_2=-\bu_2,\;\;X^2\bu_3=0. $$

Now consider the matrix

$$Y_{\bu_1}=-\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big). $$

Note that

$$ Y_{\bu_1}\bu_1=\bu_1,\;\;Y_{\bu_1}\bu_2=Y_{\bu_1}\bu_3=0. $$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i. $$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.

Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}. $$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$ A(e^{tX},1)\in \Mat_n^{1,2,3}. $$ Thus

$$ Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$ X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}. $$ Thus $X^2\oplus0\in\Mat_n^{1,2}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=-\bu_1,\;\;X^2\bu_2=-\bu_2,\;\;X^2\bu_3=0. $$

Now consider the matrix

$$Y_{\bu_1}=-\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big). $$

Note that

$$ Y_{\bu_1}\bu_1=\bu_1,\;\;X_{\bu_1}\bu_2=X_{\bu_1}\bu_3=0. $$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i. $$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.

Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}. $$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$ A(e^{tX},1)\in \Mat_n^{1,2,3}. $$ Thus

$$ Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$ X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}. $$ Thus $X^2\oplus0\in\Mat_n^{1,2}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=-\bu_1,\;\;X^2\bu_2=-\bu_2,\;\;X^2\bu_3=0. $$

Now consider the matrix

$$Y_{\bu_1}=-\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big). $$

Note that

$$ Y_{\bu_1}\bu_1=\bu_1,\;\;Y_{\bu_1}\bu_2=Y_{\bu_1}\bu_3=0. $$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i. $$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.

Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}. $$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$ A(e^{tX},1)\in \Mat_n^{1,2,3}. $$ Thus

$$ Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$ X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}. $$ Thus $X^2\oplus0\in\Mat_n^{1,2}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=\bu_1,\;\;X^2\bu_2=\bu_2,\;\;X^2\bu_3=0. $$

Now consider the matrix

$$Y_{\bu_1}=\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big). $$

Note that

$$ Y_{\bu_1}\bu_1=\bu_1,\;\;X_{\bu_1}\bu_2=X_{\bu_1}\bu_3=0. $$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i. $$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.

Here is an elementary proof but less elegant than the one indicated by @Robert Bryant. $\newcommand{\bR}{\mathbb{R}}$ $\DeclareMathOperator{\Mat}{Mat}$ $\DeclareMathOperator{\SO}{SO}$

Denote by $\Mat_n(\bR)$ the vector space of $n\times n$ matrices with real entries. Then $\SO(n)\subset \Mat_n(\bR)$ and I assume that you are asking what is the dimension of the smallest vector subspace of $\Mat_n(\bR)$ that contains $\SO(n)$. Denote by $V_n$ this subspace. We want to show that $V_n=\Mat_n(\bR)$ if $n\geq 3$.

For $i<j<k$ let $\Mat_n^{i,j,k}(\bR)$ be the subspace of $\Mat_n(\bR)$ consisting of matrices $A$ such that $$a_{pq}=0,\;\;\forall p,q\not\in\{i,j, k\}. $$

Note that the span of the union of the subspaces $\Mat_n^{i,j, k}(\bR)$ is $\Mat_n(\bR)$.

We claim that $\Mat_n^{i,j, k}(\bR)\subset V_n$. For simplicity we assume $i=1, j=2, k=3$. $\newcommand{\be}{\boldsymbol{e}}$

For $R\in \SO(3)$ let $A_R\in\SO(n)$ be the orthogonal transformation defined by $\newcommand{\bone}{\boldsymbol{1}}$ $A_R=R\oplus\bone$

Then $A(R,S):=A_R-A_S\in \Mat^{1,2,3}_n(\bR)$, $\forall R,S\in\SO(3)$.

For any skew-symmetric $3\times 3$ matrix $X$ we have $$ A(e^{tX},1)\in \Mat_n^{1,2,3}. $$ Thus

$$ Xe^{tX}\oplus 0=\frac{d}{dt} A(e^{tx},1)\in\Mat^{1,2,3}_n.$$ We deduce that

$$X\oplus 0=\frac{d}{dt}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}.$$

Thus $V_n$ contain all the skew-symmetric matrices in $\Mat^{1,2,3}_n$. Similarly

$$ X^2\oplus 0=\frac{d^2}{dt^2}\Big\vert_{t=0}A(e^{tX},1)\in\Mat_n^{1,2,3}. $$ Thus $X^2\oplus0\in\Mat_n^{1,2}\cap V_n$for any skew-symmetric $3\times 3$ matrix $X$. $\newcommand{\bu}{\boldsymbol{u}}$

For any orthornormal basis $\bu_1,\bu_2,\bu_3$ of $\bR^3$ there exists a skew-symmetric $3\times 3$ matrix $X=X_{\bu_1,\bu_2}$ such that

$$X^2\bu_1=-\bu_1,\;\;X^2\bu_2=-\bu_2,\;\;X^2\bu_3=0. $$

Now consider the matrix

$$Y_{\bu_1}=-\frac{1}{2}\Big( X_{\bu_1,\bu_2}-X_{\bu_2,\bu_3}+X_{\bu_1,\bu_3}\Big). $$

Note that

$$ Y_{\bu_1}\bu_1=\bu_1,\;\;X_{\bu_1}\bu_2=X_{\bu_1}\bu_3=0. $$

Define $Y_{\bu_2}$ and $Y_{\bu_3}$ in a similar fashion. For any $t_1,t_2,t_3\in\bR$ we have

$$(t_1Y_{\bu_1}+t_2Y_{\bu_2}+t_3Y_{\bu_3})\bu_i=t_i\bu_i. $$

Since any symmetric operator on $\bR^3$ is diagonalizable in some orthonormal basis we deduce that $V_n$ contains all the symmetric matrices in $\Mat_n^{1,2,3}$. Thus $\Mat^{1,2,3}_n\subset V_n$.