A\in L(V,W)$.
$$Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: Ax=0\}\leqslant V$$

Definition 3:
Let $E_A(\lambda_0)=\{x\in V:Ax=\lambda_0 x\}$. $E_A(\lambda_0)$ is the eigenspace of an eigenvalue $\lambda_0$. Note:
$$E_A(\lambda_0)=Ker(A-\lambda_0I).$$
$\lambda_0$ is an eigenvalue of $A$ $\iff$ $A$ is singular.

Definition 4:
Let $A\in M_n(\mathbb F)$. Polynomial $k_A(\lambda)\in\mathbb F[\lambda]$, $k_A=\det(A-\lambda_0I)$ is the characteristic polynomial of $A$.
Note: $\lambda_0$ is an eigenvalue of $A$ $\iff$ $k_A(\lambda_0)=0$,i.e., $\lambda_0$ is a root of the polynomial $k_A$.
$\deg(k_A)\leqslant n$

My attempt:
We have to prove: $$(\forall\lambda\in\mathbb F)\lambda\in\sigma\left(\Lambda_A\right)\implies\lambda\in\sigma(A)$$
$$\&$$
$$(\forall\lambda\in\mathbb F)\lambda\in\sigma(A)\implies\lambda\in\sigma\left(\Lambda_A\right)$$
i.e.
$$\sigma\left(\Lambda_A\right)\subseteq\sigma(A)\;\&\;\sigma(A)\subseteq\sigma\left(\Lambda_A\right)$$
For $\Lambda_A(T)$
$$AT=\lambda_0T\iff AT=\lambda_0 IT\iff(A-\lambda_0I)T=0$$
No matter how obvious this seems, I haven't come up to any justified step to prove both the matrix $A$ and the linear operator have to be singular. Everything seemed circular.
I know:
$$T\ne 0\nRightarrow \det(A-\lambda_0I)=0$$
because $T$ might be a zero-divisor and hence $\det T=0$.
Also, $\color{red}{\text{
A\in L(V, W)$. 
 $ $ Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: Ax=0\}\leqslant V$$ 

 Definition 3: 
 Let $ E_A(\lambda_0)=\{x\in V: Ax=\lambda_0 x\}$. $ E_A(\lambda_0)$ is the eigenspace of an eigenvalue $ \lambda_0$. Note: 
 $ $ E_A(\lambda_0)=Ker(A-\lambda_0I).$$ 
 $ \lambda_0 $ is an eigenvalue of $ A$ $ \iff$ $ A$ is remarkable. 

 Definition 4: 
 lease $ A\in M_n(\mathbb F)$. Polynomial $ k_A(\lambda)\in\mathbb F[\lambda]$, $ k_A=\det(A-\lambda_0I)$ is the characteristic polynomial of $ A$. 
 Note: $ \lambda_0 $ is an eigenvalue of $ A$ $ \iff$ $ k_A(\lambda_0)=0$,i.e. , $ \lambda_0 $ is a root of the polynomial $ k_A$. 
 $ \deg(k_A)\leqslant n$ 

 My attempt: 
 We have to test: $ $ (\forall\lambda\in\mathbb F)\lambda\in\sigma\left(\Lambda_A\right)\implies\lambda\in\sigma(A)$$ 
 $ $ \&$$ 
 $ $ (\forall\lambda\in\mathbb F)\lambda\in\sigma(A)\implies\lambda\in\sigma\left(\Lambda_A\right)$$ 
 i.e. 
 $ $ \sigma\left(\Lambda_A\right)\subseteq\sigma(A)\;\&\;\sigma(A)\subseteq\sigma\left(\Lambda_A\right)$$ 
 For $ \Lambda_A(T)$ 
 $ $ AT=\lambda_0T\iff AT=\lambda_0 IT\iff(A-\lambda_0I)T=0$$ 
 No matter how obvious this seems, I haven't come up to any justified footfall to raise both the matrix $ A$ and the linear hustler take to be singular. Everything seem circular. 
 I know: 
 $ $ T\ne 0\nRightarrow \det(A-\lambda_0I)=0$$ 
 because $ T$ might be a zero - divisor and therefore $ \det T=0$. 
 Also, $ \color{red}{\text {
A\in L(V,W)$.
$$Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: Ax=0\}\leqslant V$$

Bwfinitmon 3:
Let $E_A(\lambdx_0)=\{x\in V:Ax=\lambda_0 x\}$. $E_A(\lambda_0)$ is tye eitenspace of an eigenvauue $\lambdw_0$. Note:
$$E_A(\oambva_0)=Ker(A-\lambda_0I).$$
$\lamusa_0$ is ak eigskvaluz if $A$ $\iff$ $A$ is xingular.

Dexinition 4:
Let $A\hn M_u(\mathbb F)$. Polynomial $k_A(\lambda)\in\mathbf F[\lambca]$, $k_A=\det(A-\lambda_0I)$ is uhe chadacteristic polynomial of $A$.
Note: $\lajbda_0$ is an eigenvalur of $A$ $\iff$ $k_A(\lambda_0)=0$,i.e., $\lambfa_0$ id a root of the popynomial $k_A$.
$\eeg(k_W)\oeqslant n$

My attempt:
We have to probe: $$(\forall\lambda\in\mathbb F)\lambda\kn\sigka\left(\Lambea_Q\rihvt)\implies\lanbda\ig\sigma(A)$$
$$\&$$
$$(\forall\lambda\it\mathbb F)\lambda\in\sigms(A)\ikplues\lambda\in\sigma\left(\Lembda_A\right)$$
i.e.
$$\sigma\lest(\Lambda_A\siyht)\subseteq\sigma(A)\;\&\;\sigmq(A)\wubsejeq\sicma\lddt(\Lxmbsa_E\rifht)$$
For $\Lajbda_A(T)$
$$AT=\lajbda_0T\iff AT=\oambda_0 IT\iff(A-\lambda_0O)T=0$$
Gi matter how kbviouf ehis seems, I haven't come up to any justpfies step to prove both thw matrix $A$ and the lijear operwtor have to be singular. Everything seemed circuldr.
I kiod:
$$T\nt 0\kVighgqrgow \det(A-\lambda_0I)=0$$
because $T$ might be a zero-divisjd snc hence $\det T=0$.
Ajso, $\color{rec}{\tfxy{
A\in L(V,W)$. $$Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: Ax=0\}\leqslant 3: $E_A(\lambda_0)=\{x\in V:Ax=\lambda_0 $E_A(\lambda_0)$ is the Note: $\lambda_0$ is an of $A$ $\iff$ is singular. Definition 4: Let $A\in F)$. Polynomial $k_A(\lambda)\in\mathbb F[\lambda]$, $k_A=\det(A-\lambda_0I)$ is the characteristic polynomial of $A$. Note: $\lambda_0$ an eigenvalue of $A$ $\iff$ $k_A(\lambda_0)=0$,i.e., $\lambda_0$ is a root of the polynomial $\deg(k_A)\leqslant My We to prove: $$(\forall\lambda\in\mathbb F)\lambda\in\sigma\left(\Lambda_A\right)\implies\lambda\in\sigma(A)$$ $$\&$$ $$(\forall\lambda\in\mathbb F)\lambda\in\sigma(A)\implies\lambda\in\sigma\left(\Lambda_A\right)$$ i.e. $$\sigma\left(\Lambda_A\right)\subseteq\sigma(A)\;\&\;\sigma(A)\subseteq\sigma\left(\Lambda_A\right)$$ For $\Lambda_A(T)$ $$AT=\lambda_0T\iff AT=\lambda_0 IT\iff(A-\lambda_0I)T=0$$ No matter obvious this seems, I haven't come up to justified step to prove the matrix $A$ and the operator to be Everything circular. know: $$T\ne 0\nRightarrow because $T$ might be a zero-divisor and hence $\det T=0$. Also, $\color{red}{\text{
A\in L(V,W)$.
$$Ker A= A^{-1}\left(A\{0\}\right)=\{x\in V: ax=0\}\leqslant v$$

DefiNitIon 3:
leT $E_A(\lAmbdA_0)=\{x\in V:Ax=\lambda_0 x\}$. $e_a(\lamBda_0)$ is the eigenspace of an EigenVaLUe $\laMBdA_0$. Note:
$$e_A(\lambdA_0)=keR(a-\LamBdA_0I).$$
$\LamBdA_0$ Is An eigEnvAlue of $A$ $\Iff$ $A$ is singUlaR.

DEfinition 4:
Let $a\In m_n(\mathbb F)$. POlyNomial $k_A(\lambDa)\iN\mathbB F[\LamBDa]$, $k_A=\dEt(A-\LambdA_0I)$ is thE CharacTeristic pOlYNomial OF $A$.
Note: $\lAMBdA_0$ is aN eigenvalue of $A$ $\iff$ $K_a(\lAMbda_0)=0$,i.e., $\lambda_0$ is A root oF tHE pOLYnoMiaL $k_A$.
$\deg(k_A)\leQsLant n$

mY attempT:
we HAVE to PRove: $$(\forall\lamBda\in\mathbb f)\LamBda\in\sIgMa\lEFt(\LambDa_A\riGhT)\ImpLies\lambda\iN\sigMa(A)$$
$$\&$$
$$(\forall\Lambda\IN\mathbb f)\Lambda\iN\sigma(a)\imPliEs\laMBdA\iN\siGmA\LefT(\laMbdA_a\riGht)$$
i.e.
$$\sigMa\LeFt(\LamBda_A\RIGHT)\subSetEq\siGma(A)\;\&\;\sIgma(A)\subseteq\SigMa\leFT(\LaMbda_A\Right)$$
for $\LAmBda_A(T)$
$$aT=\lambDa_0T\ifF At=\lambda_0 IT\iff(A-\laMbda_0i)T=0$$
No matteR hoW oBviOuS this SEems, I hAveN't cOme up to Any justIFieD sTEP To Prove both the matrix $a$ aND ThE linear oPeratoR HaVe TO be singuLaR. EvErytHINg seeMed cIRcUlar.
I knoW:
$$T\ne 0\nRIGhTaRrow \det(a-\lAmbda_0I)=0$$
BeCauSe $T$ Might BE a zeRo-diviSor and heNce $\deT t=0$.
Also, $\color{red}{\tEXt{
A\in L(V,W)$.
$$Ker A= A^{ -1}\left(A \{0\} \ri ght )= \{x\ in V : Ax=0\}\leqsl a nt V $$

Definition 3:
Let$E_A( \l a mbda _ 0) =\{x\ in V:Ax = \l a m bda _0  x \}$ .$ E_ A(\la mbd a_0)$ i s the eige nsp ac e of an eige n va lue $\lamb da_ 0$. Note:
$$ E_A (\lamb da _0) = Ker(A -\l ambda _0I).$ $ 
$\lam bda_0$ is  a n  eigen v alue of $ A$  $\i ff$ $A$ is singul a r. 
Definition 4: 
Let $ A\ i nM _ n(\ mat hbb F)$. P ol ynomi a l $k_A( \ la m b d a)\ i n\mathbb F[\l ambda]$, $k _ A=\ det(A- \l amb d a_0I)$  is t he cha racteristic  pol ynomial o f $A$. Note: $ \ lambda_ 0$ isaneig enva l ue  o f $ A$ $\i f f$  $k _ A(\ lambda_0 )= 0$ ,i.e. , $\ l a m b da_0 $ i s arootof the polyno mia l $k _ A$. 
$\de g(k_A )\le qs lantn$

My  atte mp t:
We have to p rove : $$(\for all \l amb da \in\m a thbb F )\l amb da\in\s igma\le f t(\ La m b d a_ A\right)\implies\l am b d a\ in\sigma (A)$$$ $\ &$ $ 
$$(\for al l\l ambd a \ in\ma thbb F) \lambda\ in\sig m a( A) \implie s\ lambda \i n\s igm a\lef t (\La mbda_A \right)$ $
i.e . 
$$\sigma\left ( \Lambda_A\rig h t) \ s ub s eteq \si gma(A)\;\&\ ;\si g ma(A )\su b se teq \ sigma \left (\ L am b da_A\right)$$
For $ \L ambda_ A(T)$ 
$$AT=\lambda _0T\iff AT = \ l ambda_0IT\i f f( A -\lambda_0I)T= 0$$
N o matter h o w obviou s thi s seems,  I haven' t come uptoany  ju sti f i ed  step to prov e both  t he matr ix$A$ and  th e l ine arop erator ha ve to be  s in gu la r.Every t hing see me d c ir cul ar.
I know:$$T\n e 0\ nR ig h tar row \de t (A - \ lamb da _0 I)=0 $$be cause  $T$ mig ht be a  zero-div iso r  and  h en ce $\de t T=0$.
Also,  $ \color{red }{ \te xt{
A\in L(V,W)$.
$$Ker_A= A^{-1}\left(A\{0\}\right)=\{x\in_V: Ax=0\}\leqslant V$$

Definition 3:
Let_$E_A(\lambda_0)=\{x\in V:Ax=\lambda_0_x\}$._$E_A(\lambda_0)$ is_the_eigenspace of an_eigenvalue $\lambda_0$. Note:
$$E_A(\lambda_0)=Ker(A-\lambda_0I).$$
$\lambda_0$_is an eigenvalue of_$A$ $\iff$ $A$_is_singular.

Definition 4:
Let $A\in M_n(\mathbb F)$. Polynomial $k_A(\lambda)\in\mathbb F[\lambda]$, $k_A=\det(A-\lambda_0I)$ is the characteristic polynomial of_$A$.
Note:_$\lambda_0$ is_an_eigenvalue_of $A$ $\iff$ $k_A(\lambda_0)=0$,i.e., $\lambda_0$_is a root of the_polynomial $k_A$.
$\deg(k_A)\leqslant_n$

My attempt:
We have to prove: $$(\forall\lambda\in\mathbb F)\lambda\in\sigma\left(\Lambda_A\right)\implies\lambda\in\sigma(A)$$
$$\&$$
$$(\forall\lambda\in\mathbb F)\lambda\in\sigma(A)\implies\lambda\in\sigma\left(\Lambda_A\right)$$
i.e.
$$\sigma\left(\Lambda_A\right)\subseteq\sigma(A)\;\&\;\sigma(A)\subseteq\sigma\left(\Lambda_A\right)$$
For_$\Lambda_A(T)$
$$AT=\lambda_0T\iff_AT=\lambda_0 IT\iff(A-\lambda_0I)T=0$$
No matter_how obvious this seems, I haven't come up to_any justified step to prove both_the matrix $A$_and_the_linear operator have to_be singular. Everything seemed circular.
I know:
$$T\ne_0\nRightarrow \det(A-\lambda_0I)=0$$
because $T$ might be a_zero-divisor and hence $\det T=0$.
Also, $\color{red}{\text{